LIBRARY OF CONGRESS. 

Shelf .IE-4r 



UNITED STATES OF AMERICA. 



— A — 

MATHEMATICAL 

SOLUTION BOOK 

CONTAINING 

Systematic Solutions 



TO MANY OF THE 



Most Difficult Problems. 



TAKEN FROM THE LEADING AUTHORS ON ARITHMETIC AND ALGEBRA, MANY PROBLEMS 
AND SOLUTIONS FROM GEOMETRY, TRIGONOMETRY AND CALCULUS, 
MANY PROBLEMS AND SOLUTIONS FROM THE LEADING MATH- 
EMATICAL JOURNALS OF THE UNITED STATES, AND 
MANY ORIGINAL PROBLEMS AND SOLUTIONS, 



WITH 



NOTES AND EXPLANATIONS. 



BY 



B. F: FINKEL 

Member of the New Yoik Mathematical Society, and Profe 
in the Kidder Institute. 



KIBLER, COKELY & CO., Publishers, 
KmDER, Mo. 



ssor of Mathematics - ^ 

8 I 

7' 




\\ 






< 



Copyright, 1888, 

By 

B. !F. FINKEL, 

In the Office of the Librarian of Congress. 
WASHINGTON, D. C. 



Free Press Publishing House 
Leipsic, Ohio. 



. 



|A 



• 



y. 



DEDICATED 



TO MY FRIEND, 



PROF. FT. S. L.EHR, A. M M 



PRESIDENT OF 



THE OHIO NORMAL UNIVERSITY. 



PREFACE. 

This work is the outgrowth of eight years' experience in 
teaching in the Public Schools, during which time I have ob- 
served that a work presenting a systematic treatment of solutions 
to problems would be serviceable to both teachers and pupils. 

It is not intended to serve as a key to any work on mathe- 
matics ; but the object of its appearance is to present, for use in 
the schoolroom, such an accurate and logical method of solving 
problems as will best awaken the latent energies of pupils, and 
teach them to be original investigators in the various branches of 
science. 

It will not be denied by any intelligent educator that the so 
called "Short Cuts" and ''Lightning Methods" are positively in- 
jurious to beginners in mathematics. All the "whys" are cut 
out by these methods and the student robbed of the very object 
for which he is studying mathematics ; viz., the development of 
the reasoning faculty and the power to express his thoughts in a 
forcible and logical manner. By pursuing these methods, 
mathematics is made a mere memory drill and when the memory 
fails, all is lost ; whereas, it should be presented in such a way as 
to develop the memory, the imagination, and the reasoning fac- 
ulty. By following out the method pursued in this book, the 
mind will be strengthened in these three powers, besides a taste 
for neatness and a love of the beautiful will be cultivated. 

Any one who can write out systematic solutions to problems 
can resort to "Short Cuts" at pleasure ; but, on the other hand, 
let a student who has done all his work in mathematics by form- 
ulae, "Short Cuts," and "Lightning Methods" attempt to write 



vi PREFACE. 

out a systematic solution — one in which the work explains 
itself — and he will soon convince you of his inability to express 
his thoughts in a logical manner. These so-called "Short Cuts" 
should not be used at all, in the schoolroom. After pupils and 
students have been drilled on the systematic method of solving 
problems, they will be able to solve more problems by short 
methods than they could, by having been instructed in all tjhe 

: .J'S]aQrt Cu,ts" and "Lightning Methods" extant. . 

, It can not be denied that more time is given to, and more time 
wasted in the study of arithmetic in the public schools than in 

, any other branch of study ; and yet, as a rule, no better results 
are obtained in this branch than in, any other. The reason of 
this, to my mind, is apparent. Pupils are allowed to combine 
the numbers in such a way as "to get the answer" and that is all 
that is required. They are not required to tell why they do this, 
or why they do that, but, "did you get the answer?" is the 
question. The art of "ciphering" is thus developed at the 
expense of the reasoning faculty. 

The method of solving problems pursued in this book is 
often called the "Step Method." But we might, with equal pro- 
priety, call any orderly manner of doing any tningj tne "Step 
Method." There are oniy two methods of solving ' problems — a 
right method and a wrong method. That is the right method 
which takes up, in logical order, link by link, the chain of rea- 
soning and arrives at the correct result. Any other method is 
wrong and hurtful when pursued by those who are beginners in 
mathematics. 

One solution, thoroughly analyzed and criticised by a class? 
is worth more than a dozen solutions the difficulties of which are 
seen through a cloud of obscurities. 

This book can be used to a ^reat' advantage in the class- 
room — the problems at the end of each chapter affording ample 
exercise for supplementary work." 

Many of the Formulae in Mensuration tiave been obtained by 
the aid of the Calculus, the operation alone being indicated. This 



PREFACE. vii 

feature of the work will not detract any from its merits for those 
persons who do not understand the Calculus ; for those who do 
understand the Calculus it will afford an excellent drill to work 
out all the steps taken in obtaining the formulae. Many of the 
formulae can be obtained by elementary geometry and algebra. 
But the Calculus has been used for the sake of presenting the 
beauty and accuracy of that powerful instrument of mathematics. 

In cases in which the formulae lead to series, as in the case 
of the circumference of the ellipse, the rule is given for a near 
approximation. 

It has been the aim to give a solution to every problem 
presenting any thing peculiar, and to those which go the rounds of 
the country. Any which have been omitted will receive space 
in future editions of this work. The limits of this work have 
compelled me to omit much curious and valuable matter in 
Higher Mathematics. 

I have taken some problems and solutions from the School 
Visitor, published by John S. Royer ; the Mathematical Maga- 
zine, and the Mathematical Visitor, published by Artemas Mar- 
tin, A. M., Ph. D., LL. D.; and the Mathematical Messenger, 
published by G. H. Harvill, by the kind permission of these dis- 
tinguished gentlemen. 

It remains to acknowledge my indebtedness to Prof. William 
Hoover, A. M., Ph. D., of the Department of Mathematics and 
Astronomy in the Ohio University at Athens, for critically read- 
ing the manuscript of the part treating on Mensuration. 

Hoping that the work will, in a measure, meet the object 
for which it was written, 1 respectfully submit it to the use of 
my fellow teachers and co-laborers in the field of mathematics. 

Any correction or suggestion will be thankfully received by 
communicating the same to me immediately. 

The Author. 



CONTENTS. 



CHAPTER I. 

DEFINITIONS. 



Mathematics classified 



PAGE. 

. 11 



Definitions 



PAGE. 

11-14 



CHAPTER II. 

NUMERATION AND NOTATION. 



Numeration defined. 14 

French Method defined 14 

English Method defined 14 

Periods of Notation 15 



Notation defined 15 

Arabic Notation defined 15 

Roman Notation defined.. 15 

Examples 15 



Addition defined 



Subtraction defined 



Multiplication defined 



Division defined 



CHAPTER III. 

ADDITION. 
16 I Examples 17 

CHAPTER IV. 

SUBTRACTION. 

17 j Examples 18 

CHAPTER V. 

MULTIPLICATION. 
. . . ( 18 | Examples 19-20 

CHAPTER VI. 

DIVISION. 
20 | Examples 21 

CHAPTER VII. 

COMPOUND NUMBERS. 



Solutions . . 25-26 

Examples 26-27 



Definitions 22 

Time Measure 23 

Definitions in Time Measure. 23-25 

CHAPTER VIII. 

GREATEST COMMON DIVISOR. 

Divisor defined .... 28 j Greatest Com'n Divisor defined 28 

Common Divisor defined .28 | Examples 28-29 

CHAPTER IX. 

LEAST COMMON MULTIPLE. 

Multiple defined 29 I Least Common Multiple defined 29 

Common Multiple defined 29 I Examples 30 



CONTENTS. 



CHAPTER X. 

FRACTIONS. 



Definitions 31-33 

Fractions classified 31 



Solutions to Problems 
Examples 



PAGE. 

33-36 
36-37 



CHAPTER XI. 

CIRCULATING DECIMALS. 



I. Addition of Circulates 

II. Subtraction of Circulates. 



38 
39 



III Multiplication of Circulates 39 

IV Division of Circulates... 40 



CHAPTER XII. 

PERCENTAGE. 



Definitions 41 

Solutions 41-52 

II. Commission 52 

Definitions 52 

Solutions 52-54 

Examples 55 

Trade Discount ... 55 

Definitions 55 

Solutions 56-59 

Examples . 59 

Porfit and Loss 60 



III. 



IV. 



VI 



Definitions 60 

Examples 66-67 

Stocks and Bonds 67 

Definitions 67 

Solutions 68-78 

Examples 79-80 

. Insurance 80 

Definitions 80-81 

Solutions ... 81-84 

Examples 85 



CHAPTER XIII 

INTEREST. 



I. Simple Interest 86 

Definitions 86 

Solutions 86-89 

II. True Discount 89 

Definitions 89 

Solutions 89-90 

III. Bank Discount 90 

Definitions 90 



Solutions 91-92 

IV. Annual Interest 92 

Annual Interest defined. . . 92 
Solutions. 90-92 

V. Compound Interest 95 

Compound Interest defined 95 
Solutions 96-98 



CHAPTER XIV. 

ANNUITIES. 



Definitions 98 

Solutions 98-108 



Examples 109 



Solutions 



Definitions 122-124 

Solutions 125-127 



CHAPTER XV. 

MISCELLANEOUS PROBLEMS. 
110-122 

CHAPTER XVI. 

RATIO AND PROPORTION. 

Problems 127-129 



Analysis defined 129 

Solutions 129-160 



CHAPTER XVII. 

ANALYSIS. 

Problems 160-163 



CONTENTS. 



CHAPTER XVIII. 

ALLIGATION. 

I. Alligation Medial.. 164 I Solutions 

II. Alligation Alternate 164 | 

CHAPTER XIX. 

SYSTEMS OF NOTATION. 



PAGE. 

164-169 



Definitions 170 

Names of Systems .... 170 



Solutions 171-174 



CHAPTER XX. 

MENSURATION. 



Definitions 175-179 

Geometrical Magnitudes clas- 
sified . 175 

I. Parallelogram 181-183 

II. Triangles 183-187 

III. Trapezoid 187-188 

IV. Trapezium and Irregular 
Polygons 188 

V. Regular Polygons 188-190 

VI. Circles 190-193 

VII. Rectification of Plane 
Curves and Quadrature of 
Plane Surfaces 193-205 

VIII. Conic Sections 205 

Definitions 205-206 

. 1. Ellipse 206-209 

2. Parabola 209-211 

3. Hyperbola 211-214 

Higher Plane Curves 214 

1. The Cissoid Diodes.. 215-216 



IX. 



X. Plane Spiral 230 

1. Spirals of Archimedes.. 231 

2. The Reciprocal Spiral . . 231 

3. The Lituus 232 

4. The Logarithmic Spiral.. 232 
XL Mensuration of Solids 233 

1. Cylinder 236-237 

2. Cylindric Ungulas. . 237-244 

3. Pyramid and Cone. . 244-248 

4. Conical Ungulas 248-252 

XII. Sphere 252-258 

XIII. Spheroid 258 

1. The Prolate Spheroid 258-260 

2. The Oblate Spheroid 260-264 

XIV. Conoids 264 

1. The Parbolic Conoid 264-267 

2. Hyperbolic Conoid. 267-268 

XV. Quadrature and Cubature 
of Surfaces and Solids of Rev- 
olution 268 

1. Cycloid... 268-269 

2. Cissoid 269-270 

3. Spindles 270-271 

4. Parabolic Spindle... 271-272 

XVI. Regular Solids 272 

1 . Tetrahedron 273-274 

2. Octahedron 274 

3. Dodecahedron 274-275 

4. Icosahedron 275-276 

XVII. Prismatoid 276-277 

XVIII. Cylindric Rings. . . 277-279 

XIX. Miscellaneous Measure- 
ments 279 

1. Masons' and Bricklayers' 
work. . . . • 279 

2. Gauging 279-280 

3. Lumber Measure 280 

4. Grain and Hay 280-281 

XX. Solutions to Miscellaneous Problems 281-327 

Problems 328 

Geometry 334 

Biography of Prof. William Hoover . 338 

Probability Problems 344 

Biography of Dr. Artemas Martin 346 

Biography of Prof. E. B. Seitz , 350 



9. 
10. 
11. 



The Conchoid of Nicom- 

edes 216-217 

The Oval of Cassini 217 

The Lemniscate of Ber- 
noulli 218 

The Witch of Agnesi 218-219 

The Limacon 219 

The Quadratrix 220 

The Catenary 220-221 

The Tractrix 222 

The Syntractrix. 222 

Roulettes 222 

(a) Cycloids 222-225 

\b) Prolate and Curtate 

Cycloid . 225-226 

(c) Epitrochoid and Hy- 

potrochoid ..... 227-230 



CHAPTER I. 

DEFINITIONS. 

1. Mathematics (/xa6r//uaTix?), science) is that science 
which treats of quantity. 

f(l.) Arithmetic. ia. Differential. 

tl. Calculus h. Integral. 

(2.) Algebra... < *c. Calculus of Variations. 

*2. Quaternions. 

ca. Pure Geometry. 
1. Platouic Geometry., lb. Conic Sections. c\. Plane Trigon'y. 



Conic Sections, c\. Plane Ingony. 
Trigonometry.. 1'2. Analytical Trig. 
<3. Spherical " 



^(3.) Geometry.. { 2. Analytical Geometry. <3. Spherical 

3. Descriptive Geometry. 

'(1.) Mensuration. 

(2.) Surveying. 

(3.) Navigation. 
, (4.) Mechanics. 
s (5.) Astronomy. 

(6.) Optics. 

(7.) Gunnerv. 
1.(8.) &c, &c. 

2. JPuve Mathematics treats of magnitude or quantity 

Without relation to matter. 

3. Applied Mathematics treats of magnitude as subsist- 
ing in material bodies. 

4. Arithmetic (aplBjuf/Tix//, from d/JiB/Ao2, a number) is 
the science of numbers and the art of computing by them. 

5. Alf/ebva (Ar. al, the, and gcbcr. philosopher) is that 
method of mathematical computation in which letters and other 
symbols are employed. 

6. Geometry (yecofAeTpia y from yeoj}uerpeiy 7 to measure 
land, from yeac y y?] y the earth, and jueTpziv, to measure) is the 
science of position and extension. 

7. Calculus (Calculus, a pebble) is that branch of mathe- 
matics which commands by one general method, the most diffi- 
cult problems of geometry and physics. 



12 FINKEL'S SOLUTION BOOK. 

8. Differential Calculus is that branch of Calculus 
which investigates mathematical questions by measuring the re- 
lation of certain infinitely small quantities called differentials. 

9. Integral Calculus is that branch of Calculus which 
determines the functions from which a given differential has been 
derived. 

10. Calculus of Variations is that branch of calculus 
in which the laws of dependence which bind the variable quanti- 
ties together are themselves subject to change. 

11. Quaternions (quaternis, from quaterni four each, 
from quator, four) is that branch of algebra which treats of the 
relations of magnitude and position of lines or bodies in space by 
means of the quotient of two direct lines in space, considered as 
depending on a system of four geometrical elements, and as ex- 
pressed by an algebraic symbol of quadrinominal form. 

12. Platonic Geometry is that branch of geometry in 
which the argument is carried forward by a direct inspection of 
the figures themselves, delineated before the eye, or held in the 
imagination. 

13. JPure Geometry is that branch of Platonic geometry 
in which the argument may be practically tested by the aid of 
the compass and the square only. 

14. Conic Sections is that branch of Platonic geometry 
which treats of the curved lines formed by the intersection of a 
cone and a plane. 

15. Trigonometry (rpiycovov, triangle, jLisrpov, meas- 
ure) is that branch of Platonic geometry which treats of the re- 
lations of the angles and sides of triangles. 

16. Plane Trigonometry is that branch of trigonom- 
etry which treats of the relations of the angles and sides of plane 
triangles. 

.17. Analytical Trigonometry is that branch of trig- 
onometry which treats of the general properties and relations of 
trigonometrical functions. 

18. Spherical Trigonometry is that branch of trig- 
onometry which treats of the solution of spherical triangles. 

19. Analytical Geometry is that branch of geometry 
in which the properties and relations of lines and surfaces are in- 
vestigated by the aid of algebraic analysis. 

20. Descriptive Geometry is that branch of geometry 
which seeks the graphic solution of geometrical problems by 
means of projections upon auxiliary planes. 



DEFINITIONS. 13 

21. JMeriSWVatiori is that branch of applied mathematics 
which treats of the measurment of geometrical magnitudes. 

22. Surveying is that branch of applied mathematics 
which treats of the art of determining and representing distances, 
areas, and the relative position of points upon the earth's surface. 

23. Navigation is that branch of applied mathematics 
which treats of the art of conducting ships from one place to 
another. 

24. ]\£echanies is that branch of applied mathematics 
which treats of the laws of equilibrium and motion. 

25. Astronomy (aGTpovojuia, from affrpov, star and vojj.oS 
law) is that branch of applied mathematics in which mechan- 
ical principles are used to explain astronomical facts. 

26. Optics {07trinr] y from oniz y sight,) is that branch of 
applied mathematics which treats of the laws of light. 

27. Gunnery is that branch of applied mathematics which 
treats of the theory of projectiles. 

28. A Proposition is a statement of something proposed 
to be done. 

, rp, ( i. Lemma, 

i. Demonstrable. J ,'p , , (2, Corollary. 



29. Prop't'n. < 



2. Indemonstrable. 



a. Axiom. 

b. Postulate. 



30. A Demonstrable Proposition is one that can 
be proved by the aid of reason. 

31. A Theorem is a truth requiring a proof. 

32. A Lemma is a theorem demonstrated for the purpose 
of using it in the demonstration of another theorem. 

33. A Corollary is a subordinate theorem, the truth of 
which is made evident in the course of the demonstration of a 
more general theorem. 

34. A Problem is a question proposed for solution. 

35. An Indemonstrable Proposition can not be 

proved by any manner of reasoning. 

36. An Axiom is a self-evident truth. 

37. A Postulate is a proposition which states that some- 
thing can be done, and which is so evidently true as to require 
no process of reasoning to show that it is possible to be done. 



14 FINKEL'S SOLUTION BOOK. 

38. A Demonstration is the process of reasoning, prov- 
ing the truth of a proposition. 

39. A Solution of a problem is an expressed statement 
showing clearly how the result is obtained. 

40. An Operation is a process of finding, from given 
quantities, others that are known, by simply illustrating the 
solution. 

41. A Utile is a general direction for solving all problems 
of a particular kind. 

42. A Formula is the expression of a general rule or 
principle in algebraic language. 

43. A Seholium is a remark made at the close of a dis- 
cussion, and designed to call attention to some particular feature 
or features of it. 



CHAPTER II. 

NUMERATION AND NOTATION. 

1. Numeration is the art of reading numbers. 

2. There are two methods of numeration ; the French and the 
English. 

3. The French method is that in general use. In this method, 
we begin at the right hand and divide the number into periods 
of three figures each, and give a distinct name to each period. 

4. The English method is that used in Great Britain and the 
British provinces. In this method, we divide the number (if it 
consists of more than six figures) into periods of six figures each, 
and give a distinct name to each period. The following number 
illustrates the two methods ; the upper division showing how the 
number is read by the English method, and the lower division 
showing how it is read by the French method. 

4th period, 3d period, 2d period, 1st period. 
Trillions. Billions. Millions. Units. 

" "^845 678^904 325^47 434*913 



•e to '5 , d 



a>^ a».T5 a>.2 a>.2 n?-° S S V"2 
P-g P£ PC! P£ pO, ftS P« 

5*3 5§ 5g 5« gS si « 

5. The number expressed in words by the English method, 
reads thus: 



NUMERATION AND NOTATION. 15 

Eight hundred forty-five trillion, six hundred seventy-eight 
thousand nine hundred four billion, three hundred twenty-five 
thousand one hundred forty-seven million, four hundred thirty- 
four thousand nine hundred thirteen. 

Rejnark. — Use the conjunction a?zd, only in passing over 
the decimal point. It is incorrect to read 456,734 four hundred 
and fifty-six thousand, seven hundred and thirty-four. Omit the 
ands and the number will be correctly expressed in words. 

6. The following are the names of the Periods, according to 
the common, or French method: 

Sixth Period, Quadrillions. 



Seventh " Quintillions. 

Eighth " Sextillions. 

Ninth " Septillions. 

Tenth " Octillion. 



First Period, Units. 

Second " Thousands. 

Third " Millions. 

Fourth " Billions. 

Fifth " Trillions, 

Other periods in order are, Nonillions, Decillions, Undecil- 
lions, Duodecillions, Tredecilions, Quatuordecillions Quindecil- 
lions, Sexdecillions, Septendecillions, Octodecillions, Novende- 
cillions, Vigintillions, Primo-Vigintillions. Secundo-vigintillions, 
Tertio-vigintillions, Quarto- vigintillions, Quinto-vigintil lions, 
Sexto-vigintillions, Septo-vigintillions, Octo-vigintillions, Nono- 
vigintillions, Trigillions; Primo-Trigillions, Secundo-Trigillions, 
and so on to Quadragillions ; Primo-quadragillions, Secundo- 
quadragillions, and so on to Quinquagillions; Primo-quinqua- 
giilions, Secundo-quinquagillions. and so on to Sexagillions, 
Primo-sexagillions, Secundo-sexagillions. and so on to Septua- 
gillions ; Primo-septuagillions, Secundo-septuagillions, and so on 
to Octogillions ; Primo-octogillions, Secundo-octogillions. and 
so on to Nonogillions ; Primo-nonogillions, Secundo-nonogillions, 
and so to Centillions. 

7. Notation is the art of writing numbers. 

There are three methods of expressing numbers ; by words, 
by letters, called the Romci7i method, and by figures, called the 
Arabic method. 

8. The Roman Notation, so called from its having originated 
with the ancient Romans, uses seven capital letters to express 
numbers; viz., I, V, X, L, C, D, M. 

9. The Arabic Notation, so called from its having been made 
known through the Arabs, uses ten characters to express num- 
bers ; viz., 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. 

EXAMPLES. 

1. Write three hundred seventy quadrillion, one hundred one 
thousand one hundred thirty-four trillion, seven hundred eighty- 



16 FINKEL'S SOLUTION BOOK 

nine thousand six hundred thirty-two billion, two hundred ninety- 
eight thousand seven hundred sixty-five million, four hundred 
thirty-seven thousand one hundred fifty-six. 

2. Read by the English method, 78943278102345789328903- 

24678. 

3. Write three thousand one hundred forty-one quintillion, 
five hundred ninety-two billion six hundred fifty-three million 
five hundred eighty- nine thousand seven hundred ninety- three 
quadrillion, two hundred thirty -eight billion four hundred sixty- 
two million six hundred forty-three thousand three hundred 
eighty-three trillion, two hundred seventy-nine billion five hun- 
dred two million, eight hundred eighty-four thousand one hundred 
ninety-seven. 

4. Read 141421356237309504880168872420969807856971437- 
89132. 

5. Is a billion, a million million? Explain. 

6. Write 19 billion billion billion. 

7. Write 19 trillion billion million million. 

8. Write 19 hundred 56 thousand. 

9. Write 457 thousand 341 million. 

10. Write 19 trillion trillion billion billion million million. 



CHAPTER III. 

ADDITION. 



1. A.dditiOTl is the process of uniting two or more numbers 
of the same kind into one sum or amount. 

2. Add the following, beginning at the right, and prove the 
result by casting out the 9's : 

7845 excess of 9=6' 



6780 


t < 


" 9=3 


8768 


n 


» 9=2>Excess of 9's-- 


5343 


a 


" 9=6 


3987 


i i 


" 9=0 J 



32723 excess of 9=8 

Explanation. — Adding the digits in the first number, we 
have 24. Dividing by 9, we have 6 for a remainder, which is the 
excess of the 9's. Treating the remaining numbers in the same 
manner, we obtain the excesses 3, 2, 6, 0. Adding the excesses 
and taking the excess of their sum, we have 8 ; this being equal 
to the excess of the sum the work is correct. 



SUBTRACTION. 17 



Add the following, beginning at the left : 

8456 
9799 
4363 
5809 
5432 



31 
26 
23 
29 



33859 

From this operation, we see that it is more convenient to be- 
gin at the right 

Remark. — We can not add 8 apples and 5 peaches because 
we can not express the result in either denomination. Only 
numbers of the same name can be added. 

EXAMPLES. 

1. Add the numbers comprised between 20980189 and 
20980197. 

2. 6095054 + 900703+90300420+9890655+37699+29753 = 
what? 

3. Add the following, beginning at the left: 97674; 347- 
893; 789356; 98935679; 123456789. 

4. Add all the prime numbers between 1 and 107 inclusive. 

5. Add 31989, 63060, 132991, 1280340, 987654321, 78903, and 
prove the result by casting out the 9's. 

6. Add the consecutive numbers from 100 to 130. 

7. Add the numbers from 9897 to 9910 inclusive. 

8. Add MDCCCLXXVI, MDCXCVIII, DCCCCXLIX, 
DCCCLXII. 



CHAPTER IV. 

SUBTRACTION. 



1. Subtraction is the process of finding the difference be- 
tween two numbers. 

2. Subtract the following and prove the result by casting 
out the 9's : 

984895 excess of 9's=7 



795943 " " 9's=l 



I 

188952 " " 9's=6j 



Excess of 9's=7. 



18 FINKEL'S SOLUTION BOOK. 

Expla7tation. — Adding the digits in the first number, we 
have 43. Dividing by 9 the remainder is 7, which is the excess 
of the 9's. Treating the subtrahend and remainder in the same 
manner, we have the excesses 1 and 6. But subtraction is the 
opposite of addition and since the minuend is equal to the sum 
of the subtrahend and remainder, the excess of the sum of the 
excesses in the subtrahend and remainder is equal to the excess 
in the minuend. This is the same proof as that required if we 
were to add the subtrahend and remainder. 

3. We begin at the right to subtract, so that if a figure of 
the subtrahend is greater than that corresponding to it in the 
minuend, we can borrow one from the next higher denomination 
and reduce it to the required denomination and then subtract. 

4. Subtract the following and illustrate the process : 

1=9 9 9 9 9 9 9+1 1=9 9 9 9 9 9+1 ) . , , 1=9 9 9+1 ) A , , 

90000000 9856342 \ Add - 4326546 " 
85784895 8978567 3214957 



421510 5 877775 1111589 
EXAMPLES. 

1. From 9347893987 take 8968935789. Prove the result by 
casting out the 9's. 

2. 7847893578— 6759984699=what ? 

Which is the nearer number to 920864; 1816090 or 27497? 

4. 34567— 3451 8 + 3—2 + 3—4 + 7 + 18—567 + 43812 — 1326 + 
678=what. ? 

5. 5 + 6 + 7—12—13 + 14—2—3 + 7—8—6 + 5 + 12— 8=what ? 

6. 3 + 4— (6 + 7)— 8 + 27— (1 + 3— 2— 3) — (7— 8 + 5) 3 + 7= 
what ? 



CHAPTER V. 

MULTIPLICATION. 



1. Multiplication is the process of taking one number as 
many times as there are units in another; or it is a short method 
of addition when the numbers to be added are equal. 

2. Multiply the following and prove the result by casting 
out the 9's : 

7855 excess of 9's=7 
435 " " 9's=3 

39275 21 excess of 9's=3. 

23565 
31420 



3416925=excess of 9's=3, 



MULTIPLICATION. 19 

Explanation. — Adding the digits in the multiplicand and 
dividing the sum by 9, the remainder is 7 which is the excess of 
the 9's. Adding the digits in the multiplier and dividing the sum 
by 9, we have the remainder 3 which is the excess of the 9's. 
Now, since multiplication is a short method of addition when 
the numbers to be added are equal, we multiply the excess in the 
multiplicand by the excess in the multiplier and find the excess, 
and this being equal to the excess in the product, the work is 
correct. 

3. Multiply the following, beginning at the left : 

75645 
765 



1st .... 49 
35 
42 

2d 4228 

3035 
36 

3d 3524 

2530 
30 
20 
25 



57868425 



3. From this operation, we see that it is more convenient to 
begin at the right to multiply. 

5. In multiplication, the multiplicand may be abstract, or 
concrete; but the multiplier is always abstract. 

6. The sign of multiplication is X? and is read, multiplied by, 
or times. When this sign is placed between two numbers it de- 
notes that one is to be multiplied by the other. In this case, it 
has not been established which shall be the multiplicand and 
which the multiplier. Thus 8x5=40, either may be considered 
the multiplicand and the other the multiplier. If 8 is the mul- 
tiplicand, we say, 8 multiplied by 5 equals 40, but if 5 is the 
multiplicand we say, S times 5 equals 40. 

EXAMPLES. 

1. 562402 X 345728=what ? 

2. 1 mile = 63360 inches; how many inches from the earth 
to the moon the distance being 239000 miles? 

3. Multiply 789627 by 834, beginning at the left to multiply. 

4. 1 acre = 43560 sq. in.; how many square inches in a field 
containine: 427 acres? 



20 FINKEL'S SOLUTION BOOK. 

5. Multiply 6934789643 by 34789. Prove the result by cast- 
ing out the 9's. 

6. 2778588 X 34678=what ? 

7. 2X3X4— 3X7+3— 2X2+4+8X2+4— 3x5+27=what? 

8. 5 X 7+6 X 7+8 X 7—4 X 6+6 X 6+7 X 6=what ? 

9. 356789X4876= what? 

10. 395076 X 576426=what ? 

11. 7733447 X998800=what? 

12. 5654321 X999880=what? 



CHAPTER VI 

DIVISION. 



1. Division is the process of finding how many times one 
number is contained in another; or, it is a short method of sub- 
traction when the numbers to be subtracted are equal. 

2. Divide the following and prove the result by casting out 
the 9's: 

67)5484888(81864 
536 



— — Dividend 

124 5484888 excess of 9's=0. 

67 

— Quotient 

578 81864 excess of 9's=01 Excess of 9's 

536 \{ n this product 

Divisor , ~ 

428 67 excess of 9's=4J equals 0. 

402 

~268 
268 

Explanation. — Adding the digits in the dividend and di- 
viding the sum by 9, we have the remainder 0, which is the ex- 
cess of the 9's. Adding the digits in the quotient and dividing 
the sum by 9, we have the remainder 0, which is the excess of 
the 9's in the quotient. Adding the digits in the divisor and 
dividing the sum by 9, we have the remainder 4, which is the 
excess of the 9's in the divisor. Since division is the reverse of 
multiplication, the quotient corresponding to the multiplicand, 
the divisor to the multiplier, and the dividend to the product, we 
multiply the excess in the quotient by the excess in the divisor. 
The excess of this product is 0. This excess being equal to the 
excess of the 9's in the dividend, the work is correct. 



DIVISION. 21 

If there be a remainder after dividing, find its excess and add 
it to the excess of the product of the excesses of the quotient and 
divisor. Take the excess of the sum and if it is equal to the ex- 
cess of the dividend the work is correct. 

3. The sign of division is -4-, and is read divided by. 

4. When the divisor and dividend are of the same denomina- 
tion the quotient is abstract ; but when of different denomina- 
tions, the divisor is abstract and the quotient is the same as the 
dividend. Thus, 24 ct. -4- 4ct. = 6, and 24 ct. -^4 = 6 ct. 

Remark. — We begin at the left to divide, that after finding 
how many times the divisor is contained in the fewest left-hand 
figures of the dividend, if there be a remainder we can reduce it 
to the next lower denomination and find how many times the 
divisor is contained in it, and so on. 

Note. — The proof by casting out the 9's will not rectify 
errors caused by inserting or omitting a 9 or a 0, or by interchang- 
ing digits. 

EXAMPLES. 

1. 4326422~-961=what? Prove the result by casting out 
the 9's. 

2. 245379633477-i-1263=what? Prove the result by casting 
out the 9's. 

3. What number multiplied by 109 with 98 added to the 
product, will give 106700? 

4. The product of two numbers is 212492745 ; one of the 
numbers is 1035; what is the other number? 

5. 27H-9 X 3-4-9 — 1+3—3 X 9— 8-J-4+5 X 2 — 3 X 2-4-2—3 — 
(3X4-4-6+5— 2)+81-r-27x3-r-9Xl8-5-6=what? [Hint.— Per- 
form the operations indicated by the multiplication and division 
signs in the exact order of their occurrence.] 

6. (64^32x96-f-12 — 7— 5+3) X J [(27-9-3)^-9 — 1+2] + 
91-^13x7— 45 j X9+45-4-9+3— l=what? 

7. 2x2-4-2-r-2-r-2x2x2-4-2-i-2-4-2=what? Ans. \. 

8. 3^-3-4-3 X3x3-r-3-4-0x4x4x 5 X5=what? Ans. oo. 

9. 2x2x2-4-2x2-r-2-4-2x2x2x0x2X2=what? Ans. 0. 



22 FINKEL'S SOLUTION BOOK. 

CHAPTER VII. 

COMPOUND NUMBERS. 

1. A Compound Nuftlber is a number which expresses 
several different units of the same kind of quantity. 

2. A Denominate Number is a concrete number in 
which the unit is a measure; as, 5 feet, 7 pints. 

3. The Terms of a compound number are the numbers of 
its different units. Thus, in 4 bu. 3 pk. 7 qt. 1 pt., the terms 
are 4 bu. and 3 pk. and 7 qt. and 1 pt. 

4. Reduction of Compound Numbers is the process 

of changing a compound number from one denomination to an- 
other. There are Two Cases, Reduction Descending and Re- 
duction Ascending. 

5. Reduction Descending is the process of reducing a 
number from a higher to a lower denomination. 

6. Reduction Ascending is the process of reducing a 
number from a lower to a higher denomination. 

Ex. Reduce 2 E. Fr. 1 E. En. 2 E. Fl. 3 yd. 2 na. to nails. 




E. Fr. E. En. E. Sc. E. Fl. yd. 



4 
qr. 



12qr. 5qr. 



6qr. 12 qr. 
6 » 
5" 
12 " 

35qr. 
4 

140 na. 

2" 



na. 

2 



142 na. 



TIME MEASURE. 23 



TIME MEASURE. 

1. Time is a measured portion of duration. 

2. The measures of time are fixed by the rotation of the earth 
on its axis and its revolution around the sun. 

3. A Day is the time of one rotation of the earth on its 

axis. 

4. A Yeav is the time of one revolution of the earth 
around the sun. 

TABLE. 

60 seconds (sec.) make 1 minute (min.) 



60 minutes 
24 hours 
7 days 
4 weeks 
13 lunar months, 1 da. 6 hr. 
12 calender months 
365 days 

365 da". 5 hr. 48 min. 46.05 sec. 
365 da. 6 hr. 9 min. 9 sec. 

365 da. 6 hr. 13 min. 45.6 sec. 

366 days 
354 days 

19 years 

28 years 

15 years 

532 years 



1 hour (hr.) 

1 day (da.) 

1 week (wk.) 

1 lunar month (mo.) 

1 year (yr.) 

1 year. 

1 common year. 

1 solar year. 

1 sidereal year. 

1 Anomalistic year. 

1 leap year, or bissextile year. 

1 lunar year. 

1 Metonic cycle. 

1 solar cycle. 

1 Cycle of Indiction. 

1 Dionysian Period. 



5. The unit of time is the day. 

6. The Sidereal Day is the exact time of one rotation 
of the earth on its axis. It equals 23 hr. 56 min. 4.09 sec. 

7. The Sclav Day is the time between two successive 
appearances of the sun on a given meridian. 

8. The AstVOflomical Day is the solar day, begin- 
ning and ending at noon. 

9. The Civil Day, or Mean Sclav Day, is the average 
of all the solar days of the year. It equals 24 hr. 3 min. 
56.556 sec. 

10. The Sclav Year, or Ti'cpieal Yeav, is the time 
between two successive passages of the sun through the vernal 
equinox. 

11. The Sidereal Yeav is the time of a complete revolu- 
tion of the earth about the sun, measured by a fixed star. 

12. The Ancmalistic Yeav is the time of two suc- 
cessive passages of the earth through its perihelion. 

13. A Lunav Yeav is 12 lunar months and consists of 
354 day. 



24 FINKEL'S SOLUTION BOOK. 

14. A WetOflic Cycle is a period of 19 solar years, after 
which the new moons again happen on the same days of the year. 

15. A Solar Cycle is a period of 28 solar years, after 
which the first day of the year is restored to the same day 
of the week. To find the year of the cycle, we have the fol- 
lowing rule: 

Add nine to the date, divide the sum by twenty-eight; the 
quotient is the number of cycles, and the remainder is the year of 
the cycle. Should there be no remainder the proposed year is the 
twenty-eighth, or last of the cycle. The formula for the above 

rule is < f in which x denotes the date, and r the re- 

mainder which arises by dividing j-j-9 by 28, is the number 
required. 

Thus, for 1892, we have (1 892+9 )-s-28=67ff- •"• 1892 is the 
25th year of the 68 cycle. 

16. The Lunar Cycle is a period of 19 years, after which 
the new moons are restored to the same day of the civil month. 

The new moon will fall on the same days in any two years 
which occupy the same place in the cycle; hence, a table of the 
moon's phases for 19 years will serve for any year whatever 
when we know its number in the cycle. This number is called 
the Golden Number. 

To find the Golden Number : Add 1 to the date, divide the 
sum by 19; the quotient is the number of the cycle elapsed and the 
re?nainder is the Golden Number. 

The formula for the same is < > in which r is the re- 

mainder after dividing the date-f-1 by 19. It is the Golden 
Number. 

17. A Dionysian or Paschal Period is a period of 
532 year, after which the new moons again occur on the same 
day of the month and the same day of the week. It is obtained 
by multiplying a Lunar Cycle by a Solar Cycle. 

18. A Cycle of IndlCtion is a period of 15 years, at 
the end of which certain judicial acts took place under the Greek 
emperors. 

19. JEpact is a word employed in the calender to signify 
the moon's age at the beginning of the year. 

The common solar year, containing 365 days, and the lunar 
year only 354, the difference is 11 days; whence, if a new moon 
fall on the first of January in any year, the moon will be 11 days 
old on the first day of the following year, and 22 days on the 
first of the third year. The efact of these years are, therefore, 
eleven and twenty-two respectively. Another addition of eleven 



SOLUTIONS. 



25 



days would give thirty-three for the epact of the fourth year; but 
in consequence of the insertion of the intercalary month in each 
third year of the lunar cycle, this epact is reduced to three. In 
like manner the epacts of all the following- years of the cycle are 
obtained by successively adding eleven to the epact of the former 
year, and rejecting thirty as often as the sum exceeds that 
number. 

SOLUTIONS. 
Ex. 2. — Reduce 2 p. 3 pn. 1 tr. 1 hhd. 1 gal. 1 qt. to pints. 

126 




p- 


pn. 


tr. 


2 


3 


1 


6 


84 


42 



hhd. 
1 

63 



gal. 

1 
63 



252 gal. 252 gal. 42 gal. 



63 gal. 

42 
252 
252 



610 gal. 
4 



2440 qt. 



2441 
2 



4882 pt. 
Ex. 3. I. Reduce 2 bu. 3 pk. 2 qt. 1 pt. to pints 
Equation Method. 
1. 



Solution : II. 



Conclusion : 



2. 

3. 
4. 
5. 
6. 
7. 
8. 
19. 
III. 



1 bu.=4 pk. 

2 bu.=2x4 pk— 8 pk. 
8 pk.-j-o pk.=ll pk. 

I pk.=8 qt. 

II pk.=llx8qt.=88 qt. 
88 qt.+2 qt.=90 qt. 
lqt.=2 pt. 

90 qt.==90x2 pt.=180 pt. 
180 pt.+l pt.=181 pints. 
.-. 2 bu. 3 pk. 2 qt. 1 pt.=181 pints. 



26 



FINKEL'S SOLUTION BOOK. 



Solution: II. 



Ex. 5. 

wide, and 



Solut 



How 

1 ft. 8 

1. 
2. 

3. 
4. 
5. 

6. 

7. 
8. 

III. . 



HA 



Ex. 4. I. Reduce 529 pints to bushels. 
Equation method. 

fl. 2 pt=l qt. 

2. 529 pt.=529-^2=264 qt.+l pt. 

3. 8 qt.=l pk. 

4. 264 qt.=264-j-S=33 pk. 

5. 4 pk.=l bu. 

6. 33 pk.=33-j-4=8 bu.+l pk. 
Conclusion: III. .-. 529 pints=8 bu. 1 pk. 1 pt. 

many gallons will a tank 4 ft. long, 3 ft. 
in. deep contain? 

4 ft.=length, 

3 ft.=width, and 

1 ft. 8 in.=lf ft.==depth. 

4X3X1|=20 cubic ft.=contents of tank. 

1 cu. ft.— 1728 cu. in. 

20 cu. ft.=20Xl728 cu. in.=34560 cu. in. 

231 cu. in.=l gal. 

.-. 34560 cu. in.=34560-h231=149f^ gal. 

Conclusion : III. .*. The tank will contain 149yf gallons. 

(Fish's Comp. Arith., p. 126, prob. 2.) 

EXAMPLES. 

1. How many links in 46 mi. 3 fur. 5 ch. 25 links? 

2. How many acres in afield containing 1377 square chains? 

3. How many cubic inches in 29 cords of wood? 

4. In 1436 nails how many Ell English? 

5. How many miles in 3136320 inches? 

6. In 47 Sb. 2 1 3 3 13 19 gr. how many grains? 

7. Change 16 lb. 3 oz. 1 gr., Troy weight to Avoirdupois 
weight. 

8. An apothecary bought by Avoirdupois weight, 2 ft). 8 oz. of 
quinine at $2.40 per ounce, which he retailed at 20 ct. a scruple. 
What was his gain on the whole? 

9. How many seconds in a Dionysian Period? 

10. How many seconds in the month of February, 1892. 

11. How many seconds in the circumference of a wagon 
wheel? 

12. How long would it take a body to move from the earth to 
the moon, moving at the rate of 30 miles per day. 

13. If a man travels 4 miles per hour, how far can he travel in 
2 weeks and 3 days? 



EXAMPLES. 27 

14. How much may be gained by buying 2 hogsheads of mo- 
lasses, at 40 ct. per gallon, and selling it at 12 cents per quart? 

Ans. $10.08 

15. In 74726807872 seconds, how many solar years? 

Ans. 2368 years. 

16. At $4 per quintal, how many pounds of fish may be 
bought for $50.24? Ans. 1256 pounds. 

17. How many bottles of 3 pints each will it take to fill a 
hogshead? Ans. 168. 

18. What will 73 bushels of meal cost, at 2 cents per quart? 

Ans. $46.72. 

19. How many ounces of gold are equal in weight to 6 lb. of 
lead? Ans. 87^ oz. 

20. What is the difference between the weight of 42-J ft), of 
iron and 42.3751b. of gold? A?is. 52545 gr. 

21. How many bushels of corn will a vat hold that holds 
5000 gallons of water. Ans. 537^ 7 4 bu. 

22. A cellar 40 ft. long, 20 ft. wide and 8 ft. deep is half full 
of water. What will it cost to pump it out, at 6 cents a 
hogshead ? Ans. $22,797+. 

23. If a man buys 10 bu. of chestnuts at $5 a bushel, dry meas- 
ure, and sells the same at 25 cents a quart, liquid measure, how 
much does he gain? Ans. $43.09—)— gain. 

24. How many steps, 2 ft. 8 in. each, will a man take in walk- 
ing a distance of 15 miles? Ans. 29700. 

25. How many hair's width in a 40 ft. pole, if 48 hair's width 
equals 1 line? 

26. How many chests of tea, weighing 24 pounds each, at 43 
cents a pound, can be bought for $1548? Ans. 150 chests. 

27. How long will it take to count 6 million, at the rate of 80 a 
minute, counting 10 hours a day? Ans. 125 days. 

28. How long will it take to count a billion, at the rate of SO 
a minute, counting 12 hours a day? Ans. 

29. What will 15 hogsheads of beer cost, at 3 cents a pint. 

Ans. $194.10. 

30. How many shingles will it take to cover the roof of a 
building 60 ft. long and 56 ft. wide, allowing each shingle to be 4 
inches wide and 18 inches long, and to lie -^ to the weather? 

Ans. 20160. 

31. There are 9 oz. of iron in the blood of 1 man. How many 
men would furnish iron enough in their veins to make a plow- 
share weighing 22^ lbs.? Ans. 40. 



28 FINKEL'S SOLUTION BOOK. 

CHAPTER VIII. 
GREATEST COMMON DIVISOR. 

1. A. Divisor of a number is a number that will exactly 
divide it. 

2. A. Common Divisor of two or more numbers is a 
number that will exactly divide each of them. 

3. The Greatest Common Divisor, or Highest 
Common Factor, of two or more numbers is the greatest 
number that will exactly divide each of them. 

I. Find the G. C. D. of 60, 120, 150, 180. 
fl. 60=2X2X3X5. 
2. 120=2X2X2X3X5. 
II. {3. 150=2X3X5X5. 
4. 180=2X2X3X3X5. 
15. G.C.D.=2X 3X5=30. 
III. .-.. G. C. D.=30. 

Explanation. — By inspecting the factors of each number we 
observe that 2 is found in each set of factors; hence, each of the 
numbers can be divided by 2. But only once, since it is found 
only once in the factors of 150. We also observe that 3 will 
divide the numbers only once, since it occurs only once in the 
factors of 60 and 120. Also, 5 will divide them but once, since 
60, 120 and 180 contain it but once. Hence, the numbers, 60, 
120, 150, 180, being divisible by 2, 3 and 5, are divisible by their 
product, 2X3X5=30. 

I. Find the G. C. D. of 180, 1260, 1980. 

rl. 180=2X2X3X3X5. 
TT 1 2. 1260=2X2X3X3X5X7. 
)3. 1980=2X2X3X3X5X11- 
U. G. C. 0=2X2X3X3X5=180. 

III. .-. G. C. D. of 180, 1260, 1980=180. 

Explanation. — 2 being found twice in each number, they 
are each divisible by 2X2 or 4 ; also 3. being found twice in each 
number, they are each divisible by 3X3 or 9. 5 being found in 
each number, they are each divisible by 5. Hence, they are 
divisible by the product of these factors, 2X2X3X3X5=180. 

EXAMPLES. 

1. Find the G. C. D. of 78, 234, and 468. 

2. What is the G. C. D. of 36, 66, 198, 264, 600 and 720? 

3. I have three fields : the first containing 16 acres, the second 
20 acres, and the third 24 acres. What is the largest sized lots 



GREATEST COMMON DIVISOR. 29 

containing each an exact number of acres, into which the -whole 
can be divided? Ans. 4 A. lots. 

4. A farmer has 12 bu. of oats, 18 bu. of rye, 24 bu. of corn 
and 30 bu. of wheat. What are the largest bins of uniform size, 
and containing an exact number of bushels, into which the whole 
can be put, each kind by itself, and all the bins be full? 

Ans. 6 bu. bins. 

5. A has a four-sided field whose sides are 256, 292, 384, and 
400 feet respectively; what is the length of the rails used to fence 
it, if they are all of equal length and the longest that can be 
used? Ans. 4 ft. 

6. In a triangular field whose sides are 288,450, and 390 feet 
respectively, how many rails will it require to fence it, if the 
fence is 5 rails high, and what must be the length«of the rails if 
they lap over one foot? Ans. Length of rail, 7 ft. No. 940. 



CHAPTER IX. 

LEAST COMMON MULTIPLE. 

1. jL Jlfultijrfc °f a number is a number that will exactly 
contain it ; thus, 24 is a multiple of 6. 

2. A Common Multiple of two or more numbers is a 
number that will exactly contain each of them. 

3. The Least Common Multiple of two or more num- 
bers is the least number that will exactly contain each of them. 

I. Find the L. C. M. of 30, 40, 50. 

fl. 30=2X3X5. 
TT 12. 40=2X2X2X5. 
"■ 13. 50=2X5X5. 

U. L. C. M.=2X2X2X- > >X-~>X5=600. 
III. .-. L. C. M. of 30, 40, 50=000. 

Explanation. — The L. C. M. must contain 2 three times, or 
it would not contain 40; it must contain 5 twice, or it would not 
contain 50; it must contain 3 once, or it would not contain 30. 
Since all the factors of the numbers, 30, 40, 50, are contained in 
the L. C. M., it will contain each of them without a remainder. 

I. Find the L. C. M. of 2310, 210, 30, 6. 

fl. 2310=2X3X5X7X11- 
2. 210=2X:>X5X7. 
II. ^3. 30=2X3X5. 
4. 6=2X3. 

15. L. C. M.=2X3X5X7 XI 1=2310. 
III. .-. L. C. M. of 2310, 210, 30, 6=2310. 



FINKEL'S SOLUTION BOOK. 



Explanation. — 2 and 3 must be used, else the L. C. M* 
would not contain 6. 2, 3, and 5 must be used, else the L. C. M- 
would not contain 30. Hence 5 must be taken with the factors 
of 6. In like manner 7 must be taken with the factors already 
taken, else the L. C. M. would not contain 210. The factor 11 
must be taken with those already taken, else the L. C. M. would 
not contain 2310. Hence 2, 3, 5, 7, and 11 are the factors to be 
taken and their product 2310 is the L. C. M. 

I. The product of the L. C. M. of three numbers between 
1 and 100 is 6804 ; and the quotient of the L. C. M. divided by 
the G. C. D. is 84. What are the numbers? 



II. 



1. L. C. M.XG. C. D.=6804, and 

L. C. M 
2 - G. C. D . ' 4 ' 
3. .-. L. C. M.XG. C. D. 

G.C.D.x G - CD ' 



L. C. M. 



* G. C. D.~ 

(G. C. D.) 2 = 



L. C. M.x 

=6804-j-S4=81. 



9 
110 



III. 



L^C.M. 

G. C. D.=\/81 =9, by extracting the square root 

.-. L. C. M.=6804-~9=756. 

9=3 x 3. 

756=2 X2x3x3x3x7. 

3 x 3 x 2 x 2=36. 

3x3x3x2=54. 

3x3x7 =63. 

36, 54, and 63=the numbers. 



Explanation. — Since 9 is the G. C. D., each of the numbers 
contains the factors of 9. Since there are two 2's in the L. C. 
M., one of the numbers must contain these factors. In like man- 
ner one of the numbers must contain three 3's; one of them must 
also contain 7. •". We write two 3's for each of the numbers, two 
2's to any set of these 3's, and 3 and 7 with either of the remain- 
ing sets, observing that the product of the factors in any set does 
not exceed 100. If we omit 2 in step 9, the product of the fac- 
tors is 27- Hence 27, 36, 63 are numbers also satisfying the con- 
ditions of the problem. 

EXAMPLES. 

1. What is the L. C. M. of 13, 14, 28, 39, and 42? 

2. What is the L. C. M. of 6, 8, 10, 18, 20, 36, and 48? 

3. What is the L. C. M. of 18, 24, 36, 126, 20, 48, 96, 720, 
and 84? 

4. What is the smallest sum of money with which I can 

each, cows at $40 each, or 

Ans. $600. 



purchase a number of oxen at 
horses at $75 each? 



LEAST COMMON MULTIPLE. 



31 



5. Find three numbers whose L. C. M. is 840 and G. C. D. 

42. Ans. 84, 210, and 420. 

6. What three numbers between 30 and 140 having 12 for 
their G. C. D. and 2772 for their L. C. M. ? Ans. 36, 84, and 132. 

7. At noon the second, minute, and hour hands of a clock 
are together; how long after will they be together again for the 
first time? 

8. J. S. H. has 5 pieces of land; the first containing 3 A. 
2 rd. 1 p.; the second, 5 A. 3 rd. 15 p.; the third 8 A. 29 p.; 
the fourth, 12 A. 3 rd. 17 p.; and the fifth, 15 A. 31 p. Re- 
quired the largest sized house-lots, containing each an exact 
number of square rods, into which the whole may be divided. 

Ans. 1 A. 21 p. 

9. The product of the L. C. M. of three numbers by their 
G. C. D.=864, and the L. C.'M. divided by the G. C. D — 24; 
find the numbers. Ans. 12, 18, and 48. 



1. 



CHAPTER X. 

FRACTIONS. 
A Fraction is a number of the equal parts of a" unit. 



fl. As to For 



2. Fraction. 



'1. Common,, 

or 

Vulgar. 



2. Decimal. 



1. Simple. 

2. Complex. 

3. Compound. 



1. Proper. 
L2. As to Valued 2 - Improper. 



3. Mixed. 



Pure. 

Mixed. 

Circulati 



fl. Pure. 
mg, ]2. Mixed. 



3. Continued Fractions. 

3. A Common Fraction, or Vulgar Fraction, is 

one in which the unit is divided into any number of equal parts; 
and is expressed by two numbers, one written above the other, 



with a horizontal line between them, 
sixths. 



Th 



us, f expresses 



five- 



4. A Simple Fraction is a fraction having a single 
integral numerator and denominator; as, f. 

5. A Complex Fraction is a fraction whose numer- 



24 



ator, or denominator, or both, are fractional; as, — , ~, ~ 

ott Oir 5 



32 FINKEL'S SOLUTION BOOK. 

6. A ComjJOUnd Fraction is a fraction of a fraction; 
as, | of f. 

7. A Froper Fraction is a simple fraction whose 
numerator is less than its denominator; as, |. 

8. An Improper Fraction is a simple fraction whose 
numerator is greater than its denominator; as, J. 

9. A Mixed Number is a whole number and a frac- 
tion; as, 3f. 

10. A Decimal Fraction is a fraction whose denomi- 
nator is ten, or some power of ten; as, y 3 ^, T |- 7 , yf^- The de- 
nominator of a decimal is usually omitted and the point (.) is 
used to determine the value of the decimal expression. Thus, 

3 Q 2 7 f)97 

10 °> To"oo" — -U^W. 

11. A Fare Decimal is one which consists of decimal 
figures only; as, .375. 

13. A 3£ixed> Decimal is one which consists of an 
integer and a decimal; as, 5.25. 

13. A Circulating Decimal, or a Circulate, is a deci- 
mal in which one or more figures are repeated in the same order; 
as, .2121 etc. When a common fraction is in its lowest terms 
and the denominator contains factors other than 2 or powers of 
2, and 5 or powers of 5, the equivalent decimal fraction will be 

7 
circulating. Thus, -^-3^=— — - — — will, when reduced to a 

Zi X o X o 

decimal, be circulating because the denominator contains the 
factor 3. 

The repeating figure or set of figures is called a Repetend s 
and is indicated by placing a dot over the first and the last fig- 
ure repeated. 

14. A Fare Circulate is one which contains no figures 
but those which are repeated; as, .273. 

15. A 3£ixed Circulate is one which contains one or 
\nore figures before the repeating part; as, .45342. 

16. A Simple Fepetend contains but one figure; as, .3. 

17. A Compound Fepetend contains more than one 
figure; as, 354. 

18. Similar Fepetends are those which begin and end 
at the same decimal places; as, .3467? and .0358- 

19. Dissimilar Fepetends are those which begin or 
end at different decimal places ; as, .536, .835, and .3567. 



FRACTIONS. 33 

20. A Perfect ttepetetld is one which contains as many 
decimal places, less 1, as there are units in the denominator of the 
equivalent common fraction ; thus, y=.l42857- 

21. Conterminous RepetendS are those which end at 
the same decimal place; as, .4267, -3275? and .0321. 

22. Co-originous Hepetends are those which begin at 
the same decimal place ; as, .378, -5624, and 3-623. 

I. Reduce y 9 -^ to its lowest terms. 
3M- — -* 

°) 12 4- 

Explanation. — Dividing the numerator 9, by 3, without 
changing the denominator, the value of the fraction is dimin- 
ished as many times as there are units in the divisor 3. Dividing 
the denominator 12, by 3, without changing the numerator 9, the 
value of the fraction is increased as many times as there are units 
in the divisor 3. Hence, if we divide both terms by 3, the in- 
crease by dividing the denominator will be equal to the decrease 
by dividing the numerator, and the value of the fraction will 
remain unchanged. 

I. Reduce -| to a higher denomination. 

2 w 4 8 

Explanation. — Multiplying the numerator 2, by 4, without 
changing the denominator, the value of'the fraction is increased 
as many times as there are units in the multiplier 4. Multiply- 
ing the denominator 3, by 4, without changing the numerator, the 
value of the fraction is decreased as many times as there are 
units in the multiplier 4. Hence, if we multiply both terms by 
4, the increase by multiplying the numerator is equal to the de- 
crease by multiplying the denominator, and the value of the 
fraction remains unchanged. 

I. Reduce 9f to an improper fraction. 

,1. 9|=9+f 
■> l=f 



Solution : II. <J ~ 

A 7217 79 

-*■ 8 n^"8" — V 



3. 9=9Xf=V 2 - 



Conclusion: III. .-. 9|=y. 



I. Reduce f to 24ths. 

II 8 24 

3. |=5 times A=H- 



m. .-. t-H. 



34 



FINKEL'S SOLUTION BOOK, 



II. 



Reduce £ to 8ths. 



1. 4 11 
.JL f 8_!_3__i3 
■6 OI 8— g— g— g ' 



3. 



f=5 times -^=y 



III. .-. 4=^ 8 



I. 





5 °3 # 

*~8 












Reduce £ 


to 3ds. 








1. 


2 3 

T IP 

1 1 

2 2 ° 


f|= 


3 
2 

3 


3 


=H 


thirds. 


.-. 


'-3' 













II. 



III. 



Explanation. — In taking \ of f , we must divide 2 into the 
numerator, The denominator must be left unchanged; for that 
is the denomination to which the given fraction is to be reduced. 

I. Reduce -I to llths. 
1. 



II.J 



5 1 1 

S — TT' 

1 1 91 
l l n f 11 JE. ^ 

-o—o Ot TT _ TI _ n . 

4=3 times^W 5 



11 11 



III. .-. 4=-4=6f elevenths. 

O JJ 

I. Reduce 2 ^ 9 to a mixed number. 



II. 

III. 
I. 



II. 



III. 



1. 


3. 1 

3 x - 


2. 


2-9=29—3=9 1. 


. 29 

' 3 


— Q2 


Reduce -§-, -|, |- to their L. C. Denominator. 


' 1. 


L. C. D.==12. 


2. 


3 12 


3 TT" 


3. 


1 1 n f 12 4 

3 3 U1 1 2 1 2' 


4. 


2 9\s 4 8 

"3~ z A i 2 12' 


5. 


4 1 2 


4 12* 


6. 


1 1 n f 12. 3 

4 4 OI T2 TT 


7. 


3 Q V 3 9 

4 6 X T2 T2 * 


8. 


6 12 


6 TT' 


9. 


1 1 n f 1 2 2 

-g- — , ¥ or TY — T¥ . 


,10. 


5 K v/ 2 1 

6 ° X TT T2 ' 


. 2 


3 5 8 9 10 


* 3' 


4' 6 12' IT' T2* 



FRACTIONS. 35 



Reduce I, i, 


1 to 


their L. 


C. 


Denomina 


tor. 


1. 


L. 


C. D.= 


=40. 










2. 


1= 


-4 ft 

-40- 












3. 


i_ 

2~ 


_1 w 40_ 

~2 A40- 


_2 

-40- 










4. 


1= 




_4 v 40 - 

-5 A 40- 


_3 2 










— 40" 










.5. 


8 = 


-5 v 4 0- 

"8^4 0" 


_25 
— 40- 












M' 


5 2 

"§ TO - ' 


3 2 25 
4"0' 4 3"* 









II. 
Ill 

I ArM ; 

1. L. C. D.=24. 

2. 



II. 



1= 


_2 4 






~2 4- 






i= 


=ix 


24- 
2 4- 


— 1 8 
_ 24- 


5_ 

S 


=*x 


24_ 

TT- 


_20 
-21 


7_ 
8" 


=ix 


2.±_ 
2 4- 


_2 1 
— 2~¥ 




3.J-4 


4-1 


_1* 



.1 8 120 I 21 5 9 Oil 

"2 IT" 2" 4 i~ 2 T 21" Z 2T- 

III. .-. ; +H -i=2i|. 

I. Reduce J to a fraction whose numerator is 15. 



II 



1 2. f=f x|f=4i 

v 5 

TT] • 8 15 

111. . . T Tjy . 



IT. 



III. 



II. 



Rciuce §, |, I to equivalent fractions having 
least common numerators. 

fl. L. C. N.=12. 
9 1 1 2 

Z - X — T2- 

q 2 -e \y 1 2 1 2 

fi 
4 3 X yl2 12 

^- t — |ah — TT' 

4 
K 4 -*v 12 12 

. 2 3 4 J_2 12 12 

' ' U> 4> IT 18' TT' lT>* 

Subtract | from -j^-. 
1. L. C. D.=40. 

1 4 

L - l — To- 

•> 5 •"> \/ 4 2 5 

<>. g 8A4U IT' 

A 9 9 v 4 3 6 

4 • TO — TO * 4" 7 T) • 

n; . 9 5 86 2 5 11 

°' • • to 8 — 40 4"o — 4~0"' 



til .-. AHH* 

I. Multiply J by f. 



11.(2' 



1. IX 1 



8 ^ 6 8- 



7 X f=5 times ■A=ff b 
TTT • Z v r > 35 

111. . . gX e 48- 



36 FINKEL'S SOLUTION BOOK. 



I. Divide -| by f. 

x ' 8 • 7 8* 

2. f-j-i=7 times f= 3 -g 5 . 

iQ 5_i_3 1 of 35 35 1 11 

TTT • 5_i_3 1 1 1 

111. . . g-T-y Ig-j. 

Analysis to the last example : 



u.k 



rl. i is contained in 1, or -§-, 7 times. 
1.9. 

|- is contained in |, 5 times -fa times, or fj times. 



TT j 2. f is contained in 1, or ■§-, -^ of 7 times— |- times 
j 3. y is contained in L, |- of -J times=-2 7 4 times. 
U. fi 



Note. — By inverting the divisor, we find how many times it: is 
contained in 1. 

EXAMPLES. 

1. One-fifth equals how many twelfths? 

2. Reduce | , i, -|, and \ to fractions having a common de- 
nominator. 

3. Reduce 4 to a fraction whose numerator is 13. Ans. • 

117 
±J -8 

4. Reduce t to a fraction whose denominator is 11. Ans. Ji 

11" 

'5. Reduce -|, T , |, to fractions having common numerators. 

6 Add I 1 il S and - 7 - 

7. f of 8f— \ of 5=what? 

8. Multiply f by 8f. 

9. Multiply f of 9J of f by J of 17. 

191. ^2. 
10 - T^f-^-|f =what? 
15f 3| 

111. A3. 75. K 3. 

iJ - 10-9 Oil ' Q 9_Q4— Wllat ' 

1U Tl y TY 5 T"0 % 

12. f*XttX«-K 7 8 =what? ^«.-i- 

i H 

i 2- 1 



X Z 3 

1 
4 

i IT 

5 ^5 



H 



13. i=what? ^72.?. 907200. 



J~ 14 2l=what ? ^l/Z.? 2_5„J_6 

_8_ £2 

i H 

6* 



CIRCULATING DECIMALS. 37 

15. (2^X2H-f of T \) -i-(|)3- f .(7i_3i X |9 )=what? 

400000 
Ans. 



407511. 

!Q 7 A\ Fill ) Qll 1 J_ 2.8 1 

q]7 A 714 fiOJL ( * 43 • 43 v^i-^900 10 * c 

8 

^4;z.s. 1. 

18. Reduce 4 to thirds. ^4?z.y. 2f thirds. 

19. What fraction is as much larger than f as 4 is less than |-? 

Ans. \\. 

20. What is the value in the 13th example if a heavy mark be 
drawn between \ and \. Ans. If. 



CHAPTER XI. 

CIRCULATING DECIMALS. 

I. Change .63 to a common fraction. 

A. .63=.636363-f- etc., ad infinitum. 
II J 2. .636363+etc. _.63+.0063+.000063+etc., a d infinitum. 
13. This is a geometrical infinite decreasing series whose 
first term is .63 and ratio .63-r- 0063= T J ir . The sum 

of such a series is- — =.63-r-(l — T ^u) = ft = i I r- 

III. .-. .63=^ T . 

I. Reduce 1.001 to a common fraction. 

1. i.00i=1.00110011001100110011+etc, ad infinitum. 

2. 1.001=1.0011. 

3. 1.001 i =1+.0011+00000011 -f000000000011+etc, ad 
infinitum. 

J4. .00li=first term. 

5. T ^l^=.001 1^0000001 l=ratio. 

6. '.-. Sum=^^.0(ni^(l- T ^^)=.0011-WWV-»^9 
1-r 

7. .-. i:ooii=i+ T ^=«=i ¥ 4 T . 

(Rays H. A.,J>. 120, ex. S.) 

III. .-. i.OOl-1^9. 
Remark. — Since the denominator of the ratio is always ten or 
some power of ten, the numerator of the fraction resulting from 
subtracting the ratio from 1, will have as many 9's in it as there 




38 FINKEL'S SOLUTION BOOK. 

are ciphers in the denominator of the ratio. By dividing the first 
term by this fraction, its numerator becomes the denominator of 
the fraction required. Hence, a circulate may be reduced to a 
common fraction by writing for the denominator of the repetend 
as many 9's as there are figures in the repetend. Thus, .63= 

ft 1 1 9 

o 3 \_ 1.9 

10 — so- 

.034639 to a common fraction. 

1 034639 - 03iB39_34te_34 X 999+639_ 34X999+639 
1. .Ud4bd9-.Ud4 9 ^- 1()00 - _ 9 _ 9 _ v -_ 1Q00X999"' 

_ 34 X (1000—1) +639 34000—34+639 __ 34000+639—34 _ 
9990<X) ~~ 999000 ~~ 999000 ~~ 

34639— 34 _ 34605 _ 6921 _ 2307 _ _ 769 
999000 ~ 999000" - 199800 - 66600 — 22200' 

In case the circulate is mixed, we have the following rule : 

1. For the numerator, subtract that fart which precedes the 
repetend from the whole expression, both quantities being consid- 
ered as units. 

2. For the denominator, write as many 9's as there are figures 
in the repetend, and annex as many ciphers as there are decimal 
figures before each repetend. 

I. ADDITION OF CIRCULATES. 

I. Add 5.0770, .24, and 7.124943. 

fl. 5.0770 =5.0770 = 5.07707707 etc. 
II J 2. .24 = .242 = .24242424 etc. 

[3. 7.i24943= 7.1249431 = 7.12494312 etc. 
III. .-. Sum=12.44 12.4444444 etc.=12.44. 

Explanation. — The first thing, in the addition and subtrac- 
tion of circulates, is to make the circulates co-originous , i. e.. to 
make them begin at the same decimal place. That is, if one be- 
gins at (say) hundredths, make them all begin at hundredths, 
providing that each circulate has hundredths repeated. It is 
best to make them all begin with the circulate whose first re- 
peated figure is farthest from the decimal point, though any 
order after that may be taken. In the above example we have 
made them all begin at hundredths. After having made them 
all begin at hundredths, the next step is to make them con- 
terminous, i. e.) to make them all end at the same place. To 
do this, we find the L. C. M. of the numbers of figures repeated 
in each circulate, then divide the L. C. M. by the number of fig- 
ures repeated in each circulate for the number of times the figures 
as a group must be repeated. Thus, the number of figures in the 
first repetend is 3; in the second, 2; and in the third, 6. 

The L. C. M. of 3, 2, and 6 is 6. 6-=-3=2. .-. 770 must 



CIRCULATING DECIMALS. 



30 



be repeated twice, 
times. G— i— 6=1. 
I. Add .946 



II. 



III. 



IE. 



.946 = 

.248 = 

5.0770 = 

3.4884 = 

7.i24943= 



■f-2=3. .". 42 must be repeated three 
.*. 249431 must be taken once. 
248, 5.0770, 3.4884, and 7-124943. 
= .946 = .946666666666666 etc. 

= .2484 = .248484848484848 etc. 

-- 5.07707 = 5.077077077077077 etc. 
3.488448 = 3.488448844884488 etc. 
7.12494312 = 7.124943124943124 etc. 



=16.88562056205620+, 



1 6. Sum 

=16.885620. 
... Sum=16.88o620. 

II. SUBTRACTION OF CIRCULATES. 
Subtract 190.476 from 199.6428571 
T. 199.6428571 = 199.642857i4 
2. 190.476 =190 47619047 



9.16. 



Ill 


Difference=9.16. 








I 


Subtract 13.637 from 104.1. 










(1. 104.1 =104.14 =104.1414141 etc. 






II 


J 2. 13.637= 13.637= 13.63" 


"(').')7t) etc. 








13. Difference = 90.503776 




III 


.-. Difference=90.503776. 










III. MULTIPLICATION 


OF CIRCULATES. 




I. 


Multiply .07067 by .9432. 
.07067=070677 










.9432 = .9^- Multiply 


by the fraction t 


hus : 






.063609 


.070677 








.003056 


16 








.066665=product. 


.42406==.424062 

.7067 =.706770 










37)1.13083( 
111 

208 
185 

"233 

222 


3056= 

003056, be- 
cause the 
fraction is 



11 



40 FINKEL'S SOLUTION BOOK. 

I. Multiply 1.256784 by 6.42081. 
1.256784=1.2567842 
6.42081 = 6.420 T 9 T 

.02513568 = .025135685 1 Multiply by the 

.5027137 = .502713702 7 fraction thus: 

7.540705 =7.540705540 7 1.256784 2 

.001028270 — .001028270° .000 T 9 T 

8.069583198 11). 011811057 

.001028270 
Remark. — In multiplying by any number, begin sufficiently far 
beyond the last figure of the repetend, so that if there is any to 
carry it may be added to the repetends of the partial products, 
making them complete. Thus in the above example, when mul- 
tiplying by 4, we begin at 5, the second decimal place beyond 4, the 
last figure of the repetend ; and so when we multiply 4 by 4, the 
first figure of the repetend in the partial product is 7. 

IV. DIVISION OF CIRCULATES. 

Rule. — Change the terms to common fractions ; then divide as 
in division of fractions, and reduce the quotient to a repetend. 




Divide .75 by .1 

7 5 2 5 

9"~9 "3"3* 



1. .75=41=** 



.1 


1 

"9* 


2 5 
^3 


• 1_ 

• 9" 



:f£X9=4f=6.8181 etc.=6.81. 
'5-r-.i=6.8i. 

EXAMPLES. 

1. Add .87, .8, and 876. Ans. 2.644553. 

2. Add .3, .45, .45, .351, .6468, .6468, .6468, and 6468. 

Ans. 4.1766345618. 

3. Add 27.56, 5.632, 6.7, 16.356, .7i, and 6.i234. 

Ans. 63.1690670868888. 

4. Add 5.16345, 8.638i, and 3.75. 

Ans. 17.55919120847374090302. 

5. From 315.87 take 78.0378. Ans. 237.838072095497. 

6. From 16.1347 take 11.0884. Ans. 5.0462. 

7. From 18.1678 take 3.27. Ans. 14.895L 

8. From T 9 T take T 6 T . Ans. .i76470588235294i. 



PERCENTAGE AND ITS VARIOUS APPLICATIONS. 41 

9. From 5.12345 take 2.3523456. 

Ans. 2.7711055821666927777988888599994. 

10. Multiply 87.32586 by 4.37. Ans. 381.6140338. 

11. Multiply 382.347 by .0-1 Ans. 13.5169533. 

12. Multiply .9625668449197860 by .75. Ans. .72. 

13. Divide 234.6 by .7. Ans. 701.714285. 

14. Divide 13.5i69533 by 3.145. Ans. 4.297. 

15. Divide 2.370 by 4.923076. Ans. .481. 

16. Divide .36 by .25. Ans. 1.4229249011857707509881. 

17. Divide .72 by .75. Ans. .9625668449197860. 

18. 54.0678132-HS.594=what? Ans. 6.290. 

19. 4.956-r-.75=what? Ans. £.6087542. 

20. 7.714285-f-.952380==what? Ans. 8.1. 



CHAPTER XII. 

I. PERCENTAGE AND ITS VARIOUS APPLICATIONS. 

1. J?evce?XtCiye is a method of computation in which 100 
is taken as the basis of comparison. 

2. JPer cent, is from the Latin, per centum, per, by, and 
centum, a hundred. 

3. The Tet'iNS used in percentage are the Base, the Pate, 
the Percentage, and the Amount or Difference. 

4. The JBase is the number on which the percentage is 
computed. 

5. The Hate is the number of hundredths of the base which 
is to be taken. 

6. The Percentage is the result obtained by taking a cer- 
tain per cent, of the base. 

7. The Amount or Difference is the sum or difference 
of the base and percentage. 

CASE I. 

I. What is 10% of $700? 
r l. 100%=$700. 

II J 2. 1 % =^-U of $700=$7, and 
U. 10%=10 times $7=$70. 

III. .-. 10% of $700=$70. 



42 FINKEL'S SOLUTION BOOK. 



11.(2. 



I. What is 8% of $500? 
100%=$500, 

lo/ 0=T fo of $500==$5, and 
(3. 8%=8 times $5=$40. 
III. .-. 8% of $500=$40. 

I. What is f % of 800 men? 
A. 100%=800 men. 
II. J 2. \(f == T \j of 800 men=8 men, and 

I3. i%=f limes 8 men=6 men. 

III. .-. . f % of 800 men=6 men. 

I. What is 10% of 20% of $13.50? 
A. 100%=$13.50. 
f(l.K2. 1%=tw of $13.50=$.135, and 
IS. 20%=20 times $.135=$2.70. 
IM(2.) 100%=$2.70. 

1(3.) l%= T fo of $2.70=$.027, and 

1(4.) 10%=10 times $.027=$. 27=27 cents. 

III. .-. 10% of 20% of $13.50=27 cents.' 

I. A. had $1200 ; he gave 30% to a son, 20% of the remain- 
der to his daughter, and so divided the rest among four 
brothers that each after the first had $12 less than the 
preceding. How much did the last receive? 

rl. 100%=$1200, 
n J2. 1%=tt>o of $1200=$12, and 

( -')3. 30% =30 times $12=$360=son's share. 

U. $1200— $360=$840=remainder. 

1. 100% =$840, 

2. 1% =T ^ of $840=$8.40, and 
(2.)<j3. 20%>=20 times $8.40=$168=daughter's share. 

1 4. $840— $168=$672=amount divided among four 
[ brothers. 

(3.) 100%=fourth brother's share; 
(4.) 100%+$12=third brother's share. 
(5.) 100%+$24=second brother's share, and 
(6.) tOO% + $ 36==first mother's, share. 
(7.) 100%+(100%+$12)+(100%+$24)+(100%+ 
$36)=400%+$72=am'tthe four brothers rec'd. 
(8.) $672=amount the four brothers received. 
(9-) .-. 400%+$72=$672. 
( 10. ) 400 % =$672— $72=$600. 

(11.) l%=4Tro of $600=$1.50. 

(12.) 100%=100 times $1.50=$150=fourth brother's 
share. 

III. .-. The last received $150. (i?. H. A., p. 191,$rob. 25.) 



IIA 



PERCENTAGE AND ITS VARIOUS APPLICATIONS. 43 

I. What number increased by 20% of 3.5, diminished by 
12^% of 9.6, gives 3£? 



II. 



r(l.) 




100%=the number. 




r 1 - 


100% =3.5, 


(2.)< 


2. 


l%= T fo of 3.5=035, and 




u. 


20% =20 times mo=.7. 




r 1 - 


100% =9.6, 


(3.)< 


2. 


l%=: T i^ of 9.6=.096, and 




Is. 


12-J-%=12-J times .096=1.2. 


(4.) 




100% +.7—1.2=3+, 


(5.) 




100%— .5=3.5, and 


1(6.) 




100%=4, the number. 



III. .-. The number=4. (/?. i^". ^., /. ./07, /r<?£. #6.) 

CASE II. 

I. $90 is what % of $300? 
rl. $300=100%, 
II.<| 2. $l=^-Jo of 100%=i% . and 

13. $90=90 times £%=30%. 
III. ..-. $90 is 30% of $300. 
I. 750 men is what % of 12000 men? 
rl. 12000 men=100%, 
II J 2. 1 man= T 2io¥ of 100%=i4, ff %, and 

13. 750 men=750 times T ^%=6i-%. 
III. .-. 750 men is 6J% of 12000 men. 

I. A's money is 50% more than B's; then B'sishow many % 
less than A's? 

1. 100%=B.'s money. Then, 

2. 100%+50%=150%=A.'S money. 
II.s3. 150% = 100% of itself. 

4- 1 %=+> o of 100%=|% , and 
15. 50%=50 times f %=33£# . 
III. .*. B.'s money is 33 J% less than A.'s 

(7?. H. A., p. 192,prob. 11.) 

I. 30% of the whole of an article is how many % of -| of it? 

1. 100%=whole article. 

2. 66f %=t <> f 100%=f of the article. 

3. 66|%=i00% of itself. 

4. 1%=^7 of 100%=U%, and 
oof 

15. 30%=30 times H%=45%. 

Ill .*. 30% of the whole of an article is 45% of f of it. 

(R. H. A., p. 192, $ rob, 20.) 



ll.s 



II.< 



44 FINKEL'S SOLUTION BOOK. 

I. If a miller takes 4 quarts for toll from every bushel he 
grinds, what % does he take for toll? 

(1. 1 bu.= 32 qt. 
TT J2. 32qt— L00%, 
1X -1 3. 1 qt.=A of 100%— 3^% , and 

U. 4 qt.— 4 times 3-£%— 12|%. 

III. .-. He takes 12^% for toll. 

CASE ILL 

I. $20 is 5% of what sum? 

1. 100%— sum. 

2. 5%— $20, 

3. 1%— i of $20— $4, and 
.4. 100%— 100 times $4— $400. 

III. .-. $20 is 5% of $400. 

I. $24 is f% of what sum? 

1. 100%— sum. 

2. f%-$24, 

II.<3. ^%— ^of$24- $8, 

4. |%, or 1%,— 8 times $8— $64, and 
15. 100% — LOO times $64— $6400. 

III. .-. $24 is f % of $6400. 

I. I drew 48% of my funds in bank, to pay a note of $150 ; 
how much had I left? 

1. 100%— amount in bank. 

2. 48%— amount drawn out. 

3. 100%— 48%— 52%— amount left. 
U. 48%— $150, 

5. l%=is of $150— $3,125, and 

6. 52%— 52 times $3.125— $162.50— amount left. 

III. .-. $162.50— amount I had left. 

I. I pay $13 a month for board, which is 20% of my salary; 
what is my salary ? 

1. 100%— my monthly salary. 

2. 20%= $13, 

II.<3. 1%=A of $13— $.65, and 

4. 100%— 100 times $.65— $65, my monthly salary. 
.5. .*. $780—12 times $65— my yearly salary. 

III. .*. My salary— $780. (i?. H. A., p. 19^ prob. 20.) 



II. 



PERCENTAGE. 



45 



II 



I. $540 is 

1. 

2. 
3. 
4. 



CASE IV. 
% greater than what sum ? 



100%=sum. 

100%+8%=108%=sum increased 8%, and 
$540=sum increased 8% ; 
.-. 108%=$540. 
1%=^ of $540=$5, and 
100%=100 times $5=$500. 



III. 

I. 



11.^ 



III. 
I. 



.*. .$540 is 8% greater than $500. 

A sold a horse for $150 and gained 25% 
horse cost? 



what did the 



II. 



1. 100%— cost of horse. 

2. 25%=gain. 

3. 100%+25f tf =125%=selling price of horse, and 

4. $150=selling price of horse ; 

5. .-. 125%=$150, 

6. l%= T h of $150=$1.20, and 

7. 100%=100 times $1.20=$120=cost of horse. 

.-. The horse cost $120. 

I sold two horses for the same price, $150; on one I 
gained 25% and on the other I lost 25% ; what was the 
cost of each ? 

1. 100%=cost of first horse. 

2. 25%=gain. 

3. 100%+25%=125%=selling price of first horse, 

4. $150— selling price of hist horse; 

5. .-. 125% =$150, 

6. l%= T ^of$150=$1.20,and 

7. 100%=l66 times $1.20=$120=cost of first horse. 

1. 100%=cost of second horse. 

2. 25%=loss on second horse. 

3. 100%— 25%=75%=selling price of 2d horse, and 

4. $150=selling price of second horse; 

5. .-. 75%=$150, 

6. 1%= T V of $150=$2, and 

7. 100% =100 times $2=$2()()=cost of second horse. 



A. 



B. 



III. 



$120=cost of first horse, and 
$200=cost of second horse. 



I. A coat cost $32; the trimmings cost 70% less, and the 
making 50% less than the cloth; what did each cost? 



II.< 



46 FINKEL'S SOLUTION BOOK. 

' 1. 100%=cost of cloth. Then 

2. 100%— 70%=30%=cost of trimmings, and 

3. 100%— 50%=50%=cost of making. 

4. 100%+30%+50%=180%=cost of coat. 

5. $32=cost of coat; 

6. .-. 180%=$32, 

7- 1%= dro- of $32=$.1777*. 

8. 100%=100 times $.1777*=$17.77*=cost of cloth. 

9. 30%=30 times $.1777*=$5.33*=cost of trimming. 
10. 50%=50 times $.1777|=$8.88f=cost of making. 

?$17.77*=cost of cloth, 
III. .-. -j $ 5.33i=cost of trimmings, and 
1$ S.88f=cost of making. 

(7?. H. A.,J>. 196, prod. 12.) 



I. In a company of 87, the children are 37-J-% of the women, 
who are 44|-% of the men; how many of each? 

(1.) 100%=number of men. Then 
(2.) 44 : |%=number of women. 

rl. 100%=44*%, 
(3.)^2. l%=xio of 44f%=.44f%, and 

^3. 37*%=37-|- times .44f %=16f %=number of chil- 
dren in terms of the number of men. 
(4.) 100%+44f%+16f% = 1611%= number in the 

company, 
(5.) 87=number in the company ; 
(6.) .-. 161*96=87, 

(7.) l% =T A,of87=.54, 

(8.) 100%=100 times .54=54=number of men, 
(9.) 44f %=44|- times .54=24=number of women, and 
(10.) 16|%=16* times .54=18=number of children. 



54=number of men, 
III. .*. ■{ 24=number of women, and 
18=number of children. 

(R. H. A., p. 197,prob. 20.) 



•••< 



Our stock decreased 33-^%, and again 20%; then it rose 
20%, and again 33*%; we have thus lost $66; what 
was the stock at first? 



(4.) 



PERCENTAGE. 47 

100%=original stock. 
33^% —decrease. 

100%— 33i%=66f%=stock after first decrease. 

100%=66f%, 

l%=rk of 66f%=f%, and 
20%=20 times f%=13i%=second decrease. 
14. 66f %— 13^-%=53i%=stock after second decrease, 
rl. 100%=53i%, 

J2. l%= T foof53J%=.53i%, and 
I 3. 20%=20 times .53^%=10f %=first increase. 
U. 53£%+10| %= 64%=stock after first increase 
rl. 100%=64%, 

J 2. 1%=tt>o of 64%=.64%, and 
I 3. 33^-%>=33£ times ,64%=21£%=second increase. 
U- 64%-f-21-J-%=85£%=stock after second increase. 

100%— 85£%=14f %=whole loss; 

$66=whole loss; 

... 14f %=$66; 

1%==4- of $66=$4.50, and 
14| 

100%=100 times $4.50=$450=original stock. 
III. .-. $450=original stock. 

I. A brewery is worth 4% less than a tannery, and the tan- 
nery 16% more than the boat ; the owner of the boat 
has traded it for 75% of the brewery, losing thus $103 ; 
what is the tannery worth ? 

FIRST SOLUTION. 

(1.) 100%=value of the tannery. Then 
(2.) 100%?— 4%=9G%=value of the brewery. 

1. 100%=valueof the boat. Then [the boat. 

2. 100%+16%=116%=value of tannery in terms of 

3. 116%=100%, the value of tannery from step (1), 

4. l%=yk of 100%=!$%, and 

5. 100% c =100 times |f %=86&%=value of the boat 
in terms of the tannery. 

rl. 100%=96%, 
U A% l%= T i 1J of96%=.96%,and 
K n§. 75%=75 times .96%=72%=what the owner 

[ of the boat received for it. 

(5.) .*. 862%%— 72%=14^ ¥ %=vvhat the ownerof the 

boat lost in the trade. 
(6.) $103=what he lost; 
(7.) .-. 14^%=$103, 

(8.) l%=JL_of $103==$7.25, and 

(9.) 100%=100 times $7.25±=$725=value of tannery. 
III. .-. $725=value of the tannery. (R. H. A.,p. 197,prob.23.) 



(3.) 



48 



FINKEL'S SOLUTION BOOK. 



Remark. — The value of "the brewery and boat being ex- 
pressed in terms of the tannery, 75% of the brewery is also ex- 
pressed in terms of the tannery; hence, it is plain that the owner 

-2 6 9 % for 72% of the same value, losing 



of the boat has traded 
86&%— 72%, or 14& 



II. 



III. 



II. 



(2.) 
(3.) 

(4.) 
i- J 2 - 



(6.) 

(7-) 

(8.) 

(9.) 

(10.) 



(1) 



(2.) 



(3.) 



(4.) 

(5.) 
(6.) 



SECOND SOLUTION. 

100%=value of the boat. Then 
100%+16%=116%=value of the tannery. 

1. 100%=116%, 

2. l%= T io of H6%=1.16%, and 

3. 4%=4 times 1.16%=4.64%. 
116%— 4.64%=111.36%=the value of brewery in 

terms of the boat. 
100%=111.36%, 

1%=^ of 111.36%=1.1136%, and 
75%=75 times 1.1136%=83.52%=what the 

owner of the boat received for it. 
.-. 100%— 83.52%=16.48%=what he lost in the 

trade. 
$103=what he lost. 
.-. 16.48%=$103, 
1%= T 6^8 of $103=$6.25, and 
116%=116 times $6.25=$725=value of tannery. 
!725=value of the tannery. 

THIRD SOLUTION. 

100%=value of brewery. 

1. 100 %=value of tannery. Then 

2. 100%— 4%=96%=value of the tannery. 

3. .-. 96%.=100%, the value of brewery in step (1), 

4. 1%=^- of 100%=1.04i%, and 

5. 100%=100 times 1.04±%=104i%=value the tan- 
nery in terms of the brewery. 

1. 100%=value of boat. Then 

2. 100%+16%=116%=value of the tannery in 
terms of the boat. 

3. .'. 116%=104i%, the value of the tannery in step 
5 of (2), 

4. 1 %= T le of 104l%=.89Hf% , and 

5. 100%=100 times .89 jff %=89fff%=value of the 
boat in terms of the tannery, and consequently 
in terms of the brewery. 

•'• 89|ff%— 75%=14if|%=what the owner of 

the boat lost in the trade. 
$103=what the owner of the boat lost ; 
... 14139 % _ $10 3 ; 

1 %=-nrr^ of $103=$6.96, and 



(7.) 

(8.) 104i%=104i times $6.96=$725=value of tannery. 



J -^174 



II. 



MISCELLANEOUS PROBLEMS 49 

III. .-. $725=value of of the tannery. 

Remark. — In step 5 of (3), we have the value of the boat in 
terms of the tannery ; but the value of the tannery is in terms of 
the brewery: hence, the value of the boat is also in terms of the 
brewery. The owner of the boat, therefore, traded 89-}-^% for 
75% of the same value. 

MISCELLANEOUS PROBLEMS. 

I. A man sold a horse for $175, which was 12^-% less than 
the horse cost; what did the horse cost? 

1. 100%=cost of horse. 

2. 12£%=loss. 

3. 100%— 12|%=87|%=selling price. 

4. $175=selling price. 

5. .-. 87i%=$175, 

6. 1%=—- of $175=$2, and 

7. 100%=100 times $2=$200, 

III. .-. $200=cost of the horse. (7?. 3d p., p. 20J i ,prob. 5.) 

I. A miller takes for toll 6 quarts from every 5 bushels of 
wheat ground; what % does he take for toll.? 
(1. 1 bu.=32 qt. 

2. 5bu.=5 times 32 qt.=160 qt. 
II J 4. 160qt.=100%, 

4. 1 qt.= T i<5- of 100%=|%, and 

5. 6qt.=6 times f%=3f%. 

III. .-. He takes 3f % for toll. (7?. 3d p., p. 20£, prob. 11.) 

I. A farmer owning 45% of a tract of land, sold 540 acres, 
which was 60% ot what he owned; how many acres were 
there in the tract? 
(1.) 100%=number of acres in the tract. 

1. 100%=numbers of acres the farmer owned. 

2. 60%=number of acres the farmer sold. 

3. 540 acres=what he sold. 
(2.) J 4. .\ 60%=540 acres, 

5. I%=si5- °f 540 acres=9 acres, and 

6. 100 % =100 times 9 acres=900 acres=what he 
owned. 

(3.) 45%=what he owned. 

(4.) .-. 45%=900 acres, 
(5.) 1%= ¥ ^ of 900 acres=20 acres, and 

(6.) 100%=100 times 20 acres=2000=number of acres 
in the tract. 
III. .*. The tract contained 2000 acres. 

(i?. 3d p., p. 204, prob. 12.) 



(4.)- 



50 FINKEL'S SOLUTION BOOK. 

I. A, wishing to sell a cow and a horse to B, asked 150% 
more for the horse than for the cow; he then reduced 
the price of the cow 25%, and the horse 33-J%, at which 
price B took them, paying $290; what was the price of 
each ? 

(1.) 100%=asking price of the cow. Then 
(2-) 100%+150%=250%=asking price of the horse. 
(3.) 100%— 25%=75%=selling price of cow. 
100%=250%, 

1%= T ^ of 250%=2.50%, and 
33±%=33^ times 2.5%=83£%=reduction on the 
asking price of the horse. 
(5.) 250%— 83^%=166t%=selling price of the horse. 
IU (6.) 75%+166f%=24if%=selling price of both. 

( 7. ) $29Q=selling price of both. 
(8.) .". 241f %=$290, 

(9. ) 1 %=2^: of $290— $1.20, and 

(10.) 75%=75 times$$1.20=$90=selling price of the 

cow* 
(11.) 166f%=166| times $1.20=$200=selling price of 

the horse. 

f$ 90=selling price of the cow, and 
, ' , |$200==selling price of the horse. 

{Brooks' H. A., p. 2^3, frob. 18.) 



I. A mechanic contracts to supply dressed stone for a church 
for $87560, if the rough stone cost him 18 cents a cubic 
foot; but if he can get it for 16 cents a cubic foot, he 
will deduct 5% from his bill; required the number of 
cubic teet and the charge for dressing the stone. 

1. 100%=$87560. 

2. l%= T io of $87560=$875.60, and 

3. 5%=5 times $875.60=$4378=the deduction. 

4. 18/ — 16/— 2/=the deduction per cubic foot. 
ll.lh. .-. $4378=the deduction of 4378-f-.02, or 218900 cubic 

feet. Then 

6. $87560=cost of 218900 cubic feet. 

7. $.40=$87560-^218900=cost of one cubic feet. 
18. .'. $.40— $.18=$.22=cost of dressing per cubic foot. 



III. 



{218900=number of cubic feet, and 
22 cents=cost of dressing per cubic foot. 

{Brooks' H. A., p. 2^1, $rob. 21.) 



MISCELLANEOUS PROBLEMS. 51 



EXAMPLES. 

1. A merchant, having $1728 in the Union Bank, wishes to 
withdraw 15%; how much will remain? Ans. $1468.80. 

2. A Colonel whose regiment consisted of 900 men, lost 8% 
of them in battle, and 50% of the remainder by sickness; how 
many had he left? A?is. 414 men. 

3. What % of $150 is 25% of $36? Ans. 6%. 

4. What % of | of 4 off is £? Ans. 31$%. 

5. If a man owning 45% of a mill, should sell 33^% of his 
share for $450 ; what would be the value of the mill ? 

Ans. $3000. 

6. A. expends in a week $24, which exceeds by 33-^-% his 
earnings in the same time. What were his earnings? A?is. $18. 

7. Bought a carriage for $123.06, which was 16% less than I 
paid for a horse; what did I pay for the horse? Ans. $146.50. 

8. Bought a horse, buggy, and harness for $500. The horse 
cost 37-£% less than the buggy, and the harness cost 70% less 
than the horse ; what was the price of each ? 

Ans. buggy $275§f , horse $172^|, and harness $51f^. 

9. I have 20 yards of yard- wide cloth, which will shrink on 
sponging 4% in length and 5% in width; how much less than 
20 square yards will there be after sponging? Ans. 1A~| yards. 

10. A. found $5; what was his gain %? Ans. oo. 

11. The population of a city whose gain of inhabitants in 5 
years has been 25%, is 87500 ; what was it 5 years ago? 

Ans. 70000. 

12. The square root of 2 is what % of the square root of 3 ? 

Ans. ViTXl00%. 

13. A laborer had his wages twice reduced 10% ; what did 
he receive before the reduction, if he now receives $2.02-^ per 
day? Ans. $2.50. 

14. The cube root of 2985984 is what % of the square root of 
the same number? Ans. S-\%. 



15. A man sold two horses for the same price $210 ; on one he 
gained 25%, and on the other he lost 25%; how much did he 
gain, supposing the second horse cost him §- as much as the first? 

Ans. $10. 



52 FINKEL'S SOLUTION BOOK 

16. A merchant sold goods at 20% gain, but had it cost him 
$49 more he would have lost 15% by selling at the same price; 
what did the goods cost him? Ans. $119. 

17. If an article had cost 20% more, the gain would have 
been 25% less; what was the gain % ? Ans. 50%. 



II. COMMISSION. 

1. Commission is the percentage paid to an agent for the 
transaction of business. It is computed on the actual amount of 
the sale. 

2. An Agent, Factor, or Commission Merchant, 

is a person who transacts business for another. 

3. The JVet Proceeds is the sum left after the commission 
and charges have been deducted from the amount of the sales or 
collections. 

4. The Entire Cost is the sum obtained by adding the 
commission and charges to the amount of a purchase. 

I. An agent received $210 with which to buy goods ; after 
deducting his commission of 5%, what sum must he 
expend? 

1. 100%=what he must expend. 

2. 5%=his commission. 

3. 100%+5%=105%=what he receives. 
II.< 4. $210=what he receives. 

5. .-. 105%=$210, 

6. l%=-h of $210=$2, and 

7. 100%?=100 times $2=$200=what he expends. 

III. .-. $200=what he must expend. 

(R. 3d p., p. 207, prod. J,..) 

Note. — Since the agent's commission is in the $210, we must 
not take 5% of $210; for we w*ould be computing commission on 
his commission. Thus, 5% of ($200+$10)=$10+$.50. This is 
$.50 too much. 

I. An agent sold my corn, and after reserving his com- 
mission, invested the proceeds in corn at the same price; 
his commission, buying and selling was 3%, and his 
whole charge $12; for what was the corn first sold? 



COMMISSION. 



53 



II. 



III. 
I. 



HJ 



III. 



'(1.) 100%=cost of the corn. 
(2.) 3%=the commission. 

(3.) 100%— 3%=97%=net proceeds, which he invested 
in corn. 
1. 100%=cost of second lot of corn. 

2. 3%=the commission. 

3. 100%+3%=103%=entire cost of second lot of 
corn. 

4. 97%=entire cost of second lot of corn. 
(4.)<|5. .-. 103%=97%, 

6- l%= T ^of97%= T %%,and , 

7. 1Q0%=100 times T Vs%'=94 T 1 A %=cost of second 
lot of corn in terms of the first. 

8. 3%=3 times tV~3"%— ^tVV^ = commission on 
second lot. 

(5.) 3%+2 T %\%==5 T VV%==whole commission. 
(6.) $12=whole commission. 
(7.) .-. 5 T %V%=$12, 

(8.) 1%=-L, of $12=$2.06, and 

(9. ) 100%=100times $206=$206=cost of first lot of corn 
■. $206=cost of first lot of corn. {R. H. A., p. 219, prod. 10.) 

Sold cotton on commission, at 5%; invested the net pro- 
ceeds in sugar, commission, 2%; my whole commission 
was $210 ; what was the value of the cotton and sugar? 

(1.) 100%=cost of cotton. 

(2.) 5%=commission. [vested in sugar. 

(3.) 100%— 5%=95%=net proceeds, which he in- 

1. 100%=cost of sugar. 

2. 2%=commission. 
102%— entire cost of sugar. 

95%=entire cost of sugar. 
... 102%=95%, 

l%=Tfe of 95%= T %V%, and 
100%=100 times T %%%=:93-/ T %=cost of sugar in 

terms of cotton. 

2% =2 times T 9 T f2% =l|f %=commission on the 

sugar. 

5%+l|4%=6ff%=whole commission. 
$210=whole commission. 
••■ 0*i%=!f210, 

1 % = * of $210=$30.60, and 

100%=100 times $30.60=$3060=cost of cotton. 

93^%=93-5V times $30.60=$2850=cost of sugar. 
$?,060=cost of cotton, and 
$2850=cost of sugar. (R.H. A.,p.219,prob.6.) 



(4.) 
(5.) 
(6.) 

(-<■) 

(8.) 
(9.) 



54 



FINKEL'S SOLUTION BOOK. 



II. 



I. A lawyer received $11.25 for collecting a debt ; his com- 
mission being 5%; what was the amount of the debt? 
Q. 100%=amount of the debt. 
2. 5%=commission. 
4. $11.25— commission. 

4. .-. 5%=$11.25. 

5. 1%=| of $11.25=$2.25, and 

6. 100%=100 times $2.25=$225=amount of the debt. 
HI. .-. $225=amount of debt. 

(i?, 3d p., p. 207,prob. 6.) 

Charge $52.50 for collecting a debt of $525; what was the 
rate of commission? 

L. $525=100% 

>• $1=5*5 oflOO%==A%,and. 

I $52.50=52.5 times ^ r %=10%=rate of commission. 

•. 10%=rate of commission. 

My agent sold my flour at 4% commission; increasing the 
proceeds by $4.20, I ordered the purchase of wheat at 
2% commission; after which, wheat declining 3^-%, 
my 



I. 

II. 

I]J. 

I. 



II. 



(1.) 

(2.) 

(3.) 



(4-) 



whole loss was $5 ; what was the flour worth? 

100%=cost of flour. 

4'/o=commission on flour. 
100%— 4%=96%=net proceeds. 
1. 100%=cost of wheat. 



(5.) 

(6.) 

(7.) 
(8.) 
(9.) 

(10.) 

(11.) 
III. .-. $53= 



3. 



2%=commission on wheat. 
100%-j-2%=102%=entire C ost of wheat. 

96%-j-$4.20=entire cost of wheat. 
.-. 102%=96%+$4.20, 

l%=rk of (96%+$4.20)=.94 T 2 T %+$.0411if, 
100%=100 times (.94 T 2 T %+$.0411i|-)=94 T 2 T %+ 

$4.11|f=cost of wheat. 
2%=2 times (.94 1 2 T %+$.0411^)=lff%+$.08 T 4 T 

=commission on wheat. 
100^=94^ %+$4.111f, 

l%=!f%+$.04 T 2 T , and 
3^%=3i times («%+«.0±A)=^%+$.13H== 

loss on wheat. 

4% + Hf % + $.08 T 4 T + 3- 5 v/* + $.ia«=9A-%+ 

$.21if=whole loss. 
$5=whole loss. 
.-. 9J T %+$.21M=$5, or 
95 1 T%=$5-$.21|f=$4.78-, 2 T . 

1% = ^L of $4.78^=$.53, and 

100%=100 times $.53=$53. 
xost of flour. (R. H. A., p. 219, prob. 11.) 



COMMISSION. 55 



EXAMPLES. 

1. A broker in New York exchanged $4056 on Canal Bank, 
Portland, at -§%; what did he receive for his trouble? 

Ans. $25.35. 

2. A sold on commission for B 230 yards of cloth at $1.25 per 
yard, for which he received a commission of 3Jp% ; what was his 
commission and what sum did he remit? 

Ans. Commission $10.06i, and Remittance $277.43|. 

3. A sold a lot of books on commission of 20% , and remitted 
$160; for what were the books sold? Ans. $200. 

4. A lawyer charged $80 for collecting $200; what was his 
rate of commission? Ans. 40% 

5. I sent my agent $1364.76 to be invested in pork at $6 per 
bbl. after deducting his commission of 2%; how many barrels of 
pork did he buy? Ans. 223 bbl. 

6. How much money must I send my agent, so that he may 
purchase 250 bbl. of flour for me at $6.25 per bbl., if I pay him 
2|% commission? Ans. $1601.5625. 

7. If an agent's commission was $200, and his rate of com- 
mission 5% ; what amount did he invest? Ans. $4000 

8. My agent sold cattle at 10% commission, and after I in- 
creased the proceeds by $18, I ordered him to buy hogs at 20% 
commission. The hogs had declined 6J%, when he sold them at 
14f% commission. I lost in all $68; whatdidthe cattle sell for? 

Ans. $200. 

9. An agent sells flour on commission of 2%, and purchases 
goods on true commission of 3%. If he had received 3% for 
selling and 2% for buying, his whole commission would have 
been $5 more. Find the value of the goods bought. 

Ans. $10506. 

III. TRADE DISCOUNT. 

1. Trade Discount is the discount allowed in the pur- 
chas and sale of merchandise. 

2. A JAst 9 or Ke<juJ(tV JPrice 9 is an established price, as- 
sumed by the seller as a basis upon which to calculate discount. 

3. A Net Price is a fixed price from which no discount 
is allowed. 

4. TJte JJiscount is the deduction from the list, or regu- 
lar price. 



56 FINKEL'S SOLUTION BOOK. 



I. Sold 20 doz. feather dusters, giving the purchaser a dis- 
count of 10, 10 and 10% off, his discounts amounting to 
$325.20; how much was my price per dozen? 



ft) 



(4.)- 



100%— whosesale price. 
10% of 100%=10%=first discount. 
(3.) 100%— 10%=90%=first net proceeds. 
100%=90%, 

i%= ih of 90%= T V%, and 
10%=10 times T 9 ¥ %=9%=second discount. 
90% — 9%=81%=second net proceeds. 
100%=81%, 

(5.)^2. 1%= T U of 81%=AV%. and 

13. 10%=10 times 1 ^ Tr %=8.1%=third discount. 
(6.) 10%+9%+8.1%=27.1%=sum of discounts. 
(7.) $325. 20=sum of discounts. 
(8.) .-. 27.1%=$325.20, 
(9.) 1%=-2TT of $325.20=$12, and 
(10.) 100%=100 times $12=$1200=wholesale price 

of 20 dozen. 
(11.) $60=$1200-~20=wholesale price of 1 dozen. 

III. .-. $60=wholesale price per dozen. 

(R. 3d p., p. 209, prob. 5.) 

I. Bought 100 dozen stay bindings at 60 cents per dozen for 
40, 10, and 7|% off; what did I pay for them? 

(1.) 60/=list price of 1 dozen. 

(2.) $60=100 times $.60=list price of 100 dozen. 

fl. 100%=$60, 

2. l%= T io of$60=$.60, and 

3. 40% =40 times $.60=$24=first discount. 
.4. $60— $24=$36=first net proceeds. 

1. 100%=$36, 

2. l%=-rhr of $36=$.36. and 

3. 10%=10 times $.36=$3.60=second discount. 
.4. $36— $3.60=$32.40=second net proceeds. 

1. 100%=$32.40, 

2. 1 %=Tio of $32.40=$.324, and 

3. 7|%=7i times $.324=$2.43=third discount. 
U $32.40— $2.43=$29.97=cost. 

III. .-. I paid $29.97. (R. 3d p., p. 209, prob. 6.) 



IM 



(3.)- 



(4-)- 



(5.). 



I. A retail dealer buys a case of slates containing 10 dozen 
for $50. list, and "gets 50, 10, and 10% off; -paying for 
them in the usual time, he gets an additional 2%; what 
did he pay per dozen for the slates? 



II. 



lU 



(2.> 



(4.y< 



TRADE DISCOUNT. 57 

fl. 100%=$50. 

n J 2. l%=H)o of$50=$.50. 

1 ; | 3. 50%=50 times $.50=$25=nrst discount. 

U. $50— $25=$25=first net proceeds. 

100%=$25. 

l%=drc>° f $25=$.25. 
10%=10 times $.25=$2.50=second discount 
$25— $2.50=$22.50=second net proceeds. 
100%=$22.50. 

l%= T ^of$22.50=$.225. 
10%=10 times $.225=$2.25=third discount. 
$22.50— $2.25=$20.25=third net proceeds. 
100%=$20.25. 

l%= T J ir of $20.25=$.2025. 
^ (5.)<j3. 2 %=2 times $.2.025=$.405=fourth discount. 
4. $20.25— $.405=$19.845=cost of 10 dozen slates. 
15. $1.9845=$19.845-f-10=cost of 1 dozen slates. 

III. .*. $1.9845=cost of 1 dozen slates. 

(7?. 3d p., p. 209, prod. 9.) 

I. Sold a case of hats containing 3 dozen, on which I had re- 
ceived a discount of 10% and made a profit of 124-%' or 
37^/ on each hat ; what was the wholesale merchant's 
price per case? 

(1.) 374-/— profit on one hat. 

(2.) $13.50=36 times $.37£=profit on 3 dozen hats. 

(3.) 100%=wholesale merchant's price per case. 

(4.) 10%=discount. 

(5.) 100%— 10%=90%=my cost. 



cl. 100%=90%. 
(6.) 2. 1%- T i ir of90%=.9%, 



12V%=l2i times .9%=lli%==profit in terms of 

I wholesale price. 

(7.) .-. 1H%=$13.50. 

(8.) !%=nr of $13.50=$1.20. 

(9.) 100%=100 times $i.20=$120=wholesale mer- 
chant's price per case. 

III. .*. $120=wholesale merchant's price per case. 

(7?. 3d p., p. 21.2, prod. 4.) 

I. A bookseller purchased books from the publishers at 20% 
off the list; if he retail them at the list what w T ill be 
his per cent, of profit? 



58 



FINKEL'S SOLUTION BOOK. 



IU 



100%=list price. 
20%=discount. 

100%— 20%=80%=cost. 

100%=bookseller's selling price, because he sold them 

at the list price. 
.-. 100%— 80%=20%=gain. 

80%=100% of itself. 
1%=A- of 100%=li%, and 

20%=20 times l^%=25%=his gain %. 



III. .-. 25%=his % of profit. 



(R. 3d p., p. 211,prob. 1.) 



Note. — Observe that since his cost is 80%, and his gain 20%, 
we wish to know what % 20% is of 80%. It will become evi- 
dent if we suppose the list price to be (say) $400, and then pro- 
ceed to find the % of gain as in the above solution. 



11.^ 



Bought 50 gross of rubber buttons for 25, 10, and 5% off; 
disposed of the lot for $35.91, at a profit of 12% ; what 
was the list price of the buttons per gross? 



(1.) 

(2.) 
(3.) 

(4.V 



(6.> 



(6)< 

(7.) 

(8.) 

(9.) 

(10.) 

(11.) 



100%=list price. 

25% of !00%=25%=first discount. 
100%— 25%=75%=first net proceeds. 
100%=75%, 

1%=ttrj of 75%=!%, and 
10%=10 times f%=7i%. 
75%— 7|%=67|%=second net proceeds 
100%=67|%, 

^%=ih of 671%=.67|%, and 
5%=5 times .67|%=3.375%=third discount. 
67i%— 3.375%=64.125%=cost. 
'1. 100%=64.125%, 

2. 1%=.64125%, and 

3. 12%=12 times .64125%=7.695%=srain. 

.4. .-. 64.125%+7.695%=71.82%=selling price. 
$35.91=selling price. 
.-. 71.82%=$35.91, 
1%=^^ of $35.91==$.50, and 
100%=100 times $.50=$50=list price of 50 gross, 
$1 00=$50-=-50=list price of one gross. 



III. .*. $1.00=Hst price of one gross. 

(R. 3d p., p. 212,prob. 10.) 

I A dealer in notions buys 60 gross shoestrings at 70/ per 
gross, list, 50, 10, and 5% off; if he sell them at 20, 
10, and 5% off list, what will be his profit? 



TRADE DISCOUNT. 



59 



II. 



(3.)- 



(4.) 



(5.) 



(6.). 



III. 



70/=list price of one gross. 

$42=60 times $.70=Iist price of 60 gross. 

100%=$42. 

l%=yio of $42=$.42. 
50%=50 times $.42=$21=first discount. 
$42— $21=$21=first net proceeds. 
100%=$21. 

l%=Hro of$21=$.21. 
10%=10 times $.21=$2.10=second discount. 
$21— $2.10=$18.90=second net proceeds. 
100%=$18.90. 

1%=$.189. 

5%-=$.945=third discount. 
$18.90— $.945=$17.955=cost. 
100%=$42. 

1 %=TlW of $42=$.42. [count. 

20%=20 times $.42=$8.40=first conditional dis- 
$42 — $8.40=$33.60=first conditional net proceeds. 
100%=$33.60. 

1 %= T i Tr of $33.60=$.336. [discount. 

10%— 10 times $.336=$3.36=second conditional 

$33.60— $3.36=$30.24 = second conditional net 
proceeds. 

100%=$30.24. 

1%=tU of $30.24=$.3024. [discount. 

5%=5 times $.3024=$1.512=third conditional 
$30.24— $1.512=$28.728=selling price. 
.-. $28.728— $17.955=$10.773=his profit. 
5=his profit (7?. Sd p., f. 212, prod. 9.) 



EXAMPLES. 

1. Bought a case of slates containing 12 doz. for $80 list, and 
got 45, 10, and 10% off; getting an additional 2% off for prompt 
payment, what did I pay per dozen for the slates? 

Ans. $3.1752. 

2. Bought a case of hats containing 4 doz., on which I re- 
ceived a discount of 40, 20, 10, 5, and i\</ ( off If I sell them at 
$4 a piece making a profit of 20%, what is the wholesale mer- 
chant's price per case? Ans. $409 T y ft ^. 

3. If I receive a discount of 20, 10, and 5% off, and sell at a 
discount of 10, 5, and 2£% off; what is mv % of gain? 

Ans. 21|%— . 

4. A bill of goods amounted to $2400 ; 20% off being 
allowed, what was paid for the goods? Ans. $1920. 

5. Bought goods at 25, 20, 15, and 10% off If the sum of 
my discounts amounted to $162.30, what was the list price of the 
goods? * Ans. $300 



FINKEL'S SOLUTION BOOK, 



II. 



III. 



IV. PROFIT AND LOSS. . . 

1. Pvoftt and Loss are terms which denote the gain or loss 
in business transactions. 

2>, Profit is the excess of the selling price above the cost. 

3. LOSS is the excess of the cost above the selling price. 

I. A merchant reduced the price of a certain piece of cloth 
5 cents per yard, and thereby reduced his profit on the 
cloth from 10% to 8%; what was the cost of the cloth 
per yard ? 

100%=cost of cloth per yard. 
10%=his profit before reduction. 

8%=his profit after reduction. 
10%— 8%=2%=his reduction. 
5/=reduction. 
.". 2% =5/, 

l%=i of 5/=2i/, and 
100%=100 times 2^/=$2.50=cost per yard. 
$2.50=cost of cloth per yard. 

(R. 3d p., p. 211,prob. 13.) 

I. A dealer sold two horses for $150 each; on one he gained 
25% and on the other he lost 25%; how much did he 
lose in the transaction ? 

100%=cost of the first horse; 

25%=gain. 
100% + 25%=125%=selling price of first horse 
$150=selling price. 
.-..125% =$150, 

1%=t4t of $150=$1.20, and 
100%=100 times $1.20=$120=cost of first horse. 
$150 — $120=$30=gain on first horse. 
100%=cost of second horse. 
25%=loss. 
100% — 25%=75%=selling price of second horse. 
$150=sellin£ price. 
.-. 75%=$150, 

1%=^ of $150=$2, and 
100%=100 times $2=$200=cost of second horse. 
$200— $150=$50=loss on second horse. 
$50— -$30=$20=loss in the transaction. 
III. .'. He lost $20 in the transaction. 

(R. 3d p., p. 211, prod. 12.) 

I. A speculator in real estate sold a house and lot for $12000, 
which sale afford him a profit of 33 J % on the cost ; he 



IU 



r (l) 




(2) 


(3.) 


.(4) 


(5.) 


(6.) 


(7.) 


(8.) 




1. 




2, 




3. 


(9.)< 


4. 




5. 




6. 




7. 


(10.) v 


1(11.) 





PROFIT AND LOSS. 



61 



then invested the $12000 in city lots, which he was 
obliged to sell at a loss of 33i}%; how much did he lose 
by the two transactions? 



II. 



(I- 

(2.) 

(3.) 

(4.) 
(5.) 

(6.) 



100%=cost of the house and lot. 

33i%=gain. [lot. 

100%+33i%=133-j-%=selling price of house and 

$12000=selling price of the house and lot. 

... 133i%=$12000. 

1 %=T^n of $12000=$90. [lot. 

(7.) 100%=100 times $90=$9000=cost of house and 
(8.) $12000— $9000=$3000=gain on house and lot. 

rl. 100%=$12000. 
(9.U 2. l%=rb of $12000=$120. 

U. 33^%=33^ times $120=$4000=loss on city lots. 
(10.) $4000— $3000=$1000=loss by the two transac- 
tions. 
III. .'. $1000=his loss by the two transactions. 

(7?. 3d p., p. 211, prob. 15.) 
I. A dealer sold two horses for the same price; on one he 
gained 20%, and on the other he lost 20%; his whole 
loss was $25 ; what did each horse cost? 
(1.) 100%=selling price of each horse. 

1. 100%=cost offirst horse. 

2. 20%— gain on the first horse. 

3. 100%+20%=120%=selling price offirst horse. 

4. .-. 120%— 100%, from (1), 

5. 1 %= T h> ° f 100% =f % , and 

6. 100%=100 times f %=83$%=cost of first horse 
in terms of the selling price. 

7. 100%— 83J%=16| %=gain on first horse. 

1. 100%=cost of the second horse. 

2. 20%=loss on second horse. 

3. 100%— 20%=80%;=selling price of second horse. 
,~ . 4. .-. 80%=100%, from (1), 

5- 1 %=h J o of 100%=li% , and 

6. 100%=100 times l£%=125%=cost of second 
horse in terms of the selling price. 

7. 125% — 100%=25%=loss on the second horse. 
25%— 16f %=8£%= whole loss. 
$25=whole loss. 
.-. 8i%=$25, 

1% 1 of$25=$3, and 
' 8-^ [horse. 

100%=100 times $3=$300=selling price of each 

83£%=83£ times $3=$250=cost offirst horse. 

125%=125 times $3=$375=cost ofsecond horse. 



(2.) 



(3.) 



(4.) 
(5.) 
(6.) 

(7.) 

(8.) 

(9.) 

1(10.) 



62 FINKEL'S SOLUTION BOOK. 

III. 



f$250=cost of the first horse, and 
| $375=cost of second horse. 



II. 



I. What % is lost if § of cost equals f of selling price ? 

1. f of selling price=f of cost. 

2. £ of selling price=-^ of f- of cost=|- of cost. 

3. |- of selling price=4 times -f of cost=-| of cost. 

4. f=cost. 

5. -|=selling price. 

6. I— |=4=loss. 
7: |=100%. 
8. i=iofl00%=lli%,loss. 

III. .-. Loss=ll£%. 

I. Paid $125- for a horse, and traded him for another, giving 
60% additional money. For the second horse I received, 
a third and $25. I then sold the third horse for $150;' 
what was my % of profit or loss ? 

(1.) 100%=$125, 

(2.) l%=rir of $125=$1.25, and 

(3.) 60%=60 times $1.25=$75 = additional money 

paid for the second horse. 
(4.) $125+$75=$200=cost of second horse. 
(5.) $150=selling price of the third horse. 
(6.) $150-f$25=$175=selling price of second horse. 
(7.) $200— $175=$25=loss in the transaction. 

(1. $200=100%, 
(8.K2. $l=2io of 100%=i%, ana 

13. $25=25 times |%=12|%=my loss. 

III. .-. My loss is 12|%. (i?. H. A., p. 201, prod. 4,) 

I. If I buy at $4 and sell at $1, how many % do I lose? 

(1. $4=cost. 

2. $l=selling price. 

3. $4— $l=$3=loss. 

4. $4=100%. 

5. $l=iof 100%=25%. 

6. $3=3 times 25%=75%=loss. 

III. .-. 75%=loss. 

I. A and B each lost $5, which was 2£% of A's and 3£% of 
B's money ; which had the most, and how much? 



II. 



II. 



PROFIT AND LOSS 



63 



(1.) 100%=A's money. 

(2.) 2^%=what he lost. 

(3.) $5=what he lost. 

(4.) .-. 2|%=$5, 

(5.) 1< f> i =^L of $5=$1.80, and 

(6.) 100%=100 times $1.80=$180=A's money. 

fl. 100%=B's money. 

2. 3£%=what he lost. 

3. $5=what he lost. 
(7.)<|4. .-. 3J%=$5, 

5. 1%=5T of $5=$1.50, and 

6. 100%=100 times $1.50=$150=B , s money. 
(8.) $180— $150==$30=excess of A's money over B's. 

III. .-. A had $30 more than B. (R. H. A., p. 203, prod. 5.) 



Mr. A bought a horse and carriage, paying twice as much 
for the horse as for the carriage. He afterward sold the 
horse for 25% more than he gave for it, and the carriage 
for 20% less than he gave for it, receiving $577.50; what 
was the cost of each? 



II. 



(1.) 

(20 
(3.) 

(4.) 
(5.)- 



(6.) 

(7.) 

(8.) 

(9.) 

(10.) 

l(H.) 



100%=cost of the carriage. 
200%=cost of the horse. 

20%=loss on the carriage. 
100%— 20%=80%=selling price of the carriage. 
100% =200%, 

l%=yfo of 200%=2%, and 

25%=25 times 2%=50%=gain on the horse. 
200%+50%=250%=selling price of the horse. 

80%-|-250%=330%=selling price of both. 
$577.50=selling price of both. 
.-.330% =$57 7. 50, 

1%=^ of $577.50=$1.75, and 
100%=100 times $1.75=$175=cost of carriage. 
200%=200 times $1.75=$350=costof the horse. 



III.. 



{$175=cost of the carriage, and 
$350: 



:cost of the horse. 



( Milne's prac, p. 259, prob. 19. ) 



I. Mr. A. sold a horse for $198, which was 10% less than he 
asked for him, and his asking price was 10% more than 
the horse cost him. What did the horse cost him? 



II. 



II. 



II. 



64 FINKEL'S SOLUTION BOOK. 

(1.) 100%=cost of the horse. 

(2.) 100%+10%=110%=asking price. 

rl. 100%=110%, 
(3.K2. l%= T *o of 110%=1 T V%, and [asking price. 

13. 10%=10 times 1 T V%= 11 % = reduction from 

(4.) 110%— ll%=99%=selling price. 

(5.) $198=selling price. 

(6.) .-. 99%=$198, 

(7.) 1%=A- of $198=$2, and 

(8.) 100%=100 times $2=$200=cost of the horse. 

III. .;. $200=cost of horse. (Milne's prac, p. 259, firob. 23.) 

I. What must be asked for apples which cost me $3 per bbl., 
that I may reduce my asking price 20% and still gain 
20% on the cost? 

(1.) 100%=$3. 

(2.) 1%—rrro of $3=$.03, and 

(3.) 20%=20 times $.03=$.60=gain. 

(4.) $3.00+$.60=$3.60=selling price. 

1. 100%=asking price. 

2. 20%=reduction. 

3. 100%— 20%=80%=selling price. 
(5.)<]4. $3.60=selling price. 

5. .-. 80%=$3.60, 

6. 1%= ¥ V of $3.60=$.045, and 

7. 100%==100 times $.045=$4.50=asking price. 

III. .-. $4.50=asking price. (Milne's firac, $. 261, firob.38.) 

I. A merchant sold a quantity of goods at a gain of 20%. 
If, however, he had purchased them for $60 less than 
he did, his gain would have been 25%. What did the 
goods cost him ? 

(1.) 100%=actual cost of goods. 

(2.) 20%=gain. 

(3.) 100%+20%=120%=actual selling price. 

(4.) 100%— $60=supposed cost. 

\1. l()0%=100%— $60, 

,, J 2. 1%=^ of (100%— $60)=1%— $.60, and 

w |3. 25%=25 times (1%— $.60)= 25% —$15 = sup- 

^> posed gain. [ual selling price. 

(6.) (100%— $60) + (25% — $15)= 125%— $75=act- 

(7.) .;. 125%— $75=120%, 

(8.) 5%=$75, 

(9.) l%=iof $75=$15, and 

(10.) 100%=100 times $15=$1500=cost of the goods. 

III. .-. $1500=cost of goods. (Milne's j>rac.,j>. 261, frob. j.0.) 



(5.)- 



PROFIT AND LOSS. 65 

Note. — The selling price is the same in the last condition of 
this problem as in the first. Hence we have the selling price in 
the last condition equal to the selling price in the first as shown 
in step (7.) 

I. I sold an article at 20% gain, had it cost me $300 more, I 
would have lost 20%; find the cost. 

(1.) 100%=actual cost of the article. 

(2.) 20%=actual gain. 

(3.) 100%+20%=i20%=actual selling price. 
(4.) 100%+$300=supposed C ost. 
100%=100%+$300, 

1%=t?to of (100%+$300)=l%+$3> and 
20% =20 times ( 1 %+$3)=20%+$60=supposed 
loss. [ual selling price. 

(6.) (100%+$300)— (20%+$60)=80%+$240=act- 
(7.) .-. 120%=80%+$240. 
(8.) 40%=$240, 

(9.) 1%==A of $240=$6, and 

(10.) 100%=100 times $6=$600=cost of the article. 

III. .-. $600=cost of the article. 

(R. H. A., p. JfD9,prob. 85.) 

I. A man wishing to sell a horse and a cow, asked three 
times as much for the horse as for the cow, but, rinding 
no purchaser, he reduced the priee of the horse 20% , 
and the price of the cow 10%, and sold them for $165. 
What did he get for each? 

f (1.) 100%=asking price of the cow. 

(2.) 300%=abking price of the horse. 
(3.) 10%=reduction on the price of the cow. 

(4.) 100%— 10%=90%=selling price of the cow. 

( i. ioo%=30oy;, 

(5. ) J 2. 1 %= T io of 300%=3% , and 

(3. 20%=20 times 3%=60%=reduction on horse. 
IU (6.) 300%— 60%=240%=sellfng price ot the horse. 
(7. ) 90% ,+240%=33O# =selling price of both. 

(8.) $165=selling price of both 
(9.) .-. 330%=$165, 
(10.) l%=irk of $165=$.50, and 

(11.) 90%=90 times $.50=$45=selling price of cow. 

(12.) 240%=240 times $.50=$120=selling price of 
horse. 

Ill • > 't^ ; ">= arnoiin t ne received for the cow, and 
" , J$120=amount he received for the horse. 



66 FINKEL'S SOLUTION BOOK, 



EXAMPLES. 



1. What price must a man ask for a horse that cost him $200, 
that he may fall 20% on his asking price and still gain 20% ? 

Aits. $300. 

2. A man paid $150 for a horse which he offered in trade at 
a price he was willing to discount at 40% for cash, as he would 
then gain 20%. What was his trading price? Ans. $300. 

3. A man gained 20%? by selling his house for $3360. What 
did it cost him ? Ans. $3000. 

4. A gained 120% by selling sugar at 8/ per pound. What 
did the sugar cost him per pound? Aits. o^j^. 

5 How must cloth, costing $3-50 a yard, be marked that a 
merchant may deduct 15% from the marked price and still gain 
15%? Aits. $4.73 T 9 T . 

6. Sold a piece of carpeting for $240, and lost 20%; what 
selling price would have given me a gain of 20% ? 

Ans. $360. 

7. Sold two carriages for $240 apiece, and gained 20% on 
one and lost 20% on the other; how much did I gain or lose in 
the transaction ? Ans. Lost $20. 

8. Sold goods at a gain of 25% and investing the proceeds, 
sold at a loss of 25% ; what was my % of gain or loss. Ans. 3J%. 

9. A man sold a horse and carriage for $597, gaining by the 
sale, 25% on the horse and 10% on the cost of the carriage. If 
f of the cost of the horse equals § of the the cost of carriage, 
what was the cost of each? Ans. Carriage $270; horse $240. 

10. If |- of the selling price is gain, what is the profit? 

Ans. 80%. 

11. If -J of an article be sold for the cost of \ of it, what is 
the rate of loss? Aits. 33£%. 

12. I sold two houses for the same sum; on one I pained 25% 
and on the other I lost 25%. My whole loss was $240; what 
did each house cost? . Ans. First $1440, second $2400. 

13. My tailor informs me that it will take 10^ sq. yd. of 
cloth to make me a full suit of clothes. The cloth I am about to 
buy is 1J yards wide and on sponging it will shrink 5% in length 
and width. How many yards will it take for my new suit? 

Aits. 6 T f |s yd. 

14. A grocer buys coffee at 15/ per Bb. to the amount of $90 
worth, and sells it at the same price by Troy weight ; find the % 
of gain or loss. Aits. Gain 2 !£■§-% . 



STOCKS AND BONDS. 67 

15. I spent $260 for apples at $1.30 per bushel ; after retain- 
ing a part for my own use, I sold the rest at a profit of 40%, 
clearing $3 on the whole cost. How many bushels did I buy? 

Ans. 50 bu. 

16 How must cloth costing $3.50 per yard, be marked that 
the merchant may deduct 15% from the marked price and still 
make 15% profit? Ans. $4,735. 

17. 1 sold goods at a gain of 20%. If they had cost me $250 
more than they did, I would have lost 20% by the sale* How 
much did the goods cost me? Ans. $500. 

18. A merchant bought cloth at $3.25 per yard, and after 
keeping it 6 months sold it at $3.75 per yard. What was his 
gain %, reckoning 6% per annum for the use of money? 

Ans. 12%+. 



V. STOCKS AND BONDS. 

1. Stocks is a general term applied to bonds, state and 
national, and to certificates of stocks belong to corporations. 

3. A Dond is a written or printed obligation, under seal, 
securing the payment of a certain sum of money at or before a 
specified time. 

3. Stock is the capital of the corporation invested in busi- 
ness; and is divided into Shares, usually of $100 each. 

4. Am Assessment is a sum of money required of the 
stockholders in proportion to their amount of stock. 

5. A Dividend is a sum of money to be paid to the stock- 
holders in proportion to their amounts of stock. 

6. The Par Value of money, stocks, drafts, etc., is the 
nominal value on their face. 

'7. The JMLarket Value is the sum for which they sell. 

8. DiSCOWVit is the excess of the par value of money, 
stocks, drafts, etc., over their market value. 

9. Premium is the excess of their market value over their 
par value. 

10. DroJx'erage is the sum paid an agent for buying stocks, 
bonds, etc. 



II. 



68 FINKEL'S SOLUTION BOOK. 

I. At \°lo brokerage, a broker received $10 for making an in- 
vestment in bank stock ; how many shares did he buy ? 

1. 100 %=par value of stock. 

2. :|%=brokerage. 

3. $10=brokerage. 

4. .-. ±%=$10, 

5. 1%=4 times $10=$40, and 

6. 100%=100 times $40=$4000=par value of stock. 

7. $100=par value of one share. 

8. $4000=par value of 4000-r-lOO, or 40 shares. 

III. •. 40=number of shares, 

I. How many shares of railroad stock at 4% premium can 
be bought for $9360? 

1. 100%=par value of stock I can buy. 

2. 4%=premium. 

3. 104%=price of what I buv. 

4. $9360=price of what I buy. 
II.<!5. .-. 104%=$9360. 

6- 1 %=rk of $9360=$90. 

7. 100%=100 times $90=$9000=par value. 

8. $100=par value of one share. 
19. $9000=par value of 9000-f-100, or 90 shares. 

III. .*. 90=number of shares that can be bought. 

I. When gold is at 105, what is the value of a gold dollar in 
currency ? 

(1. 105/ ; or 105% in currency=100/; or 100% in gold. 
11.12. 1/; or 1% in currency=.95'2 5 i/ 5 or -95-^% in gold. 

1.3. 100/; or 100% in currency =9 5^/; or 95^% in gold. 
III. .'. $1 in currency is worth 95^1:/ in gold. 

I. In 1864, the "greenback" dollar was worth only 35-f-/ in 
gold; what was the price of gold? 

'l. 35f/; or 35f% in gold=100/ ; or 100% in currency. 

2. 1/; or 1% in gold=— -of 100/; or 100%=2.8/; or 
2.8% in currency. 6br ? 

3. 100/; or 100% in gold=100 times 2.8/; or 2.8%=280/; 
or 280% in currency. 

III. .*. $1 in gold was worth $2.80 in currencv. 

(R. 3d J>., p. 217,prob. 8.) 

I. Bought stock at 10% discount, which rose to 5% premium 
and sold for cash. Paying a debt of $33, I invested the 
balance in stock at 2% premium, which at par, left me 
$11 less than at first; how much money had I at first? 



II. 



STOCKS AND BONDS. 



HA 



(1.) 

(2.) 

(3.) 
(4.) 
(5.) 

(6.) 
(7.) 



(8.){ 



(90 



(10.) 

(11.) 

(12.) 
(13.) 

1(^0 



III. 



100%=my money at first. 
100%=par value of stock. 

10%=discount. 
100%— 10%=90%=market value. 
.'. 90%=100%, my money; because that is the 

amount invested. 

1%=A- of 100%=li%, and 
100%=100 times li%=lll£%=par value of the 

stock in terms of my money. 
100%=llli% 

l%=l-l-%, and [terms ot my money. 

5%=5 times li%=5-|%=premium on stock in 
llli%+5f%=116|% = what I received for the 

stock. 
116f% — $33=amount invested in second stock. 

1. 100%— par value of second stock. 

2. 2%=premium. 

3. 100%+2%=102%=marketvalueof second stock. 

4. 116f%— $33=market value of second stock 

5. .-. 102%=116f%— $33, 

6. l%= T h of (116f%-$33) = l T %%-$.32 T <y, 

7. 100%=100 times (l T ^%_$.32 T fi T )=114 f W%- 
$32 T 6 T =par value of second stock. 

114fV 8 3%— $32 T 6 T =what I received for the second 
stock, since I sold them at par. 

••• U4 T 5 A%— $32 T V=100%— $11, by the last con- 
dition of the problem. 

14 T 5 A%=$21 T <y, 

1%_1 of $21 1 fi T =$1.485, and 
100%=100 times $1.485=$148.50. 
I had $148.50 at first. (R. II. A., p. 212, f rob. 8.) 



what did 



Il.i 



III. 
I. 



I. Bought $8000 in gold at 110%, brokerage ^% 
I pay for the gold in currency? 

1. 100%=par value of gold. 

2. 110%=market value. 

3. -J%=brokerage. 

4. 110%+i%==1104%=entire cost. 

5. 100%=$8000, 

6. 1%= T ¥<t of $800O=$80, and 

7. 110-J%=H0-J times $80=$8810=cost of gold in currency. 

.'. $8000 in gold costs $8810 in currency. 

What income in currency would a man receive by invest- 
ing $5220 in U. S. 5-20, 6% bonds at 116%, when gold 
is worth 105 ? 



II. 



70 FINKEL'S SOLUTION BOOK. 

(1.) 100%=par value of the bonds. 

(2.) 116 %=market value. 

(3.) #5220=market value. 

(4.) /. 116%=$5220. 

(5.) 1%=tt6 of $5220=$45. 

(6.) 100%=100 times $45=$4500=par value of bonds. 

r l. 100%=$450C. 

(7.)<| 2. l%= T £o of $4500=145. 

U. 6%=6 times $45=$270=income in gold. 

(8.) $1.00 in gold=$1.05 in currencv. 

(9. ) $270 in gold=270 times $1.05=$283.50 in currency. 

III. .*. $283.50=income in currency. 

(R. 3dp.,p.217,prob. 5.) 



I. What % of income do U. S. 4-J% bonds, at 108, yield when 
gold is 105%? 

(1.) 100%=amount invested in the bonds. 

(2.) 100%=par value of bonds. 

(3.) 108%=market value. • 

(4.) .-. 108%=100%, from (1). 

(5.) 1%=^ of 100%=-|f%, [of amount invested. 

(6.) 100%=100 times §f %=92| %=par value in terms 

rl. 100%=92i%. 

(7.)<2. l%= T io of 92i%=|+%. 

13. 4£%=4i times ff %=4i%=income in gold. 
(8.) 100% in gold=105% in currency. 
(9.) 1% in gold= T i-o of 105%=l2-V% in currency. 

(10.) 4£% in gold=4| times l^%=4f % in currency. 



II. 



III. .'. Income in currency=4|-%. 

Note. — This is a general solution of the preceding problem. 
Since there is no special amount given, we represent the amount 
invested by 100%. The market value and the amount invested 
being the same, we have 108%=100% as shown in (4). 

I. A man bought Michigan Central at 120, and sold at 124% ; 
what % of the investment did he gain? 

1. 124%=selling price. 

2. 120%=cost. 

3. 124%— 120%=4%=gain. 

4. 120%=100% of itself. 

5. l%= T ioOf 100%=f%, 

6. 4%=4 times |%=3^%=gain on the investment. 

III. .-. He gained 3-J-% on the investment, 



II. 



STOCKS AND BONDS. 



71 



I. What sum invested in U. S. 5's of 1881, at 118, yielded an 
annual income of $1921 in currency, when gold was 
at 113? 



$1.13 in currency=$l in gold. 

$1 in currency= T 1 T 3 of $l==$J-f|in gold, and 

$1921 in currency=1921 times' $|™==$1700==in- 

come in gold. 
100%=par value of the bonds. 

5%=income in gold. 
$1700=income in gold. 
.-. 5%=$1700, 

\%=\ of $1700=$340, and [bonds. 

100%=i00 times $340=$34000=par value of the 
100%=$34000, 

l%=TTnr of $34000=$340, and 
118%=118 times $340=$40120=market value, or 

amount invested. 



llA 




III. .'. $40120=amount invested. 



SECOND SOLUTION. 



II.< 



III. 



I. 



(I-) 

(2.) 
(3.) 

(4.) 
(5.) 
(6.) 



100%=amount invested in currency. 
100%=par value. 
118%=market value. 
.-. 118%=100%, from (1.) 

1%=TT8 of 100%=4£%, and 
100%=100 times 4«%=84±±#=par value in 
terms of the investment. 
rl. 100%=84f|%. 
(7.)<2. l%=rbof84|f%=M%^nd 

13. 5%=5 times \% %=4£|%=income in gold, 
rl. 100% in gold=113% in currency, 

1% in gold=l3*'(^% in currency, and 
41f% in gold— 4iA times l T ^-%=4 T ^y/*=income 

in currency. 
$1921=income in currency. 
•'• 4^V%=$1921, 

1 %=j4r of $1921=4401.20, and 

100%=100 times $401.20=$40120=amount in- 
vested in currency. 

. $40120=amount invested. (R. 3d p., p. 218, frob. 8.) 



(8.) 

(9.) 

(10.) 

(11.) 
(12.) 



How many shares of stock bought at 95]% , and sold at 
105, brokerage \ c / c on each transaction, will yield an 
income of $925? 



72 



FINKEL'S SOLUTION BOOK. 



II.S 



1. 

2. 
3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 

11. 

12. 
13. 
14. 



100%=par value of stock. 
95£%=market value of stock. 

^%=brokerage. 
95i+i%=95i%=entire cost. 
105%=selling price-|-brokerage. 



XG 



%=brokerage. 
105%— i%=104f %=selling price. 
104|%— 95|%==9i%=gain. 
$925=gain. 



9i' 



,=$925. 
1 



1%=~ of $925=$100, and 

100%=100 times $100=$10000=par value of stock. 

$1.00=par value one share. 

$10000=par value 10000—100, or 100 shares. 



III. .'. 100=number of shares. 



(R. 3d p., p. 218,prob. 9.) 



I. 



If I invest all my money in 5% furnace stock salable at 
75%, my income will be $180; how much must I bor- 
row to make an investment in 5% state stock selling at 
102%, to have that income? 

1. 100%=par value of furnace stock. 

2. 5%=income. 

3. $180=income. 

4. .-. 5%=$180, 

5. l%=i of $180=$36, and [nace stock. 

6. 100%=100 times $36=$3600=par value of fur- 
fl. 100%=$3600, 
' 2. l%= T fo ot $3600=$36, and [nace stock. 

75%z=75 times $36=$2700=market value of fur- 
100%=par value of state stock. 

6%=income. 
$180=income. 
.-. 6%=$180, 

1 %=l of $180=$30, and [stock. 

100%=100 times $30=$3000=par value of state 
100% =$3000, 

1% =T ^ of $3000=$30, and [state stock. 

102%?=102 times $30=.$3060=market value of 
$3060— $2700=$360=what I must borrow. 




III. .-. I must borrow $360. 



(R. H. A., p. 225, prob. 2.) 



When U. S. 4% bonds are quoted at 106, what yearly in- 
come will be received in gold from bonds that can be 
bought for $4982? 



STOCKS AND BONDS. 



73 



II. 



III. 
I. 



(1.) 100%=par value of the bonds. 

(2.) 106 %=market value. 

(3.) $4982=market value, or amount invested. 

(4.) .-". 106%=$4982, 

(5.) l%=rk of $4982=$47, and 

(6.) 100%=100 times $47=$4700. 

A. 100%=$4700, 
(!.){% l%= T io of $4700=$47, and 

13. 4%=4 times $47=$188=income in gold. 
.. $188=income in gold. (7?. 3p., p. 218, prod. 11.) 

The sale of my farm cost me $500, but I gave the pro- 
ceeds to a broker, allowing him ^%, to purchase rail- 
road stock then in the market at 102%; the farm paid 
5% income, equal to $2075, but the stock will pay 
$2025 more; what is the rate of dividend? 



II. 



(10 

(2.) 
(30 

(40 
(5.) 

(60 
(70 



(»0<i 



6. 



100'%=value of the farm. 

5%=income on the farm. 
$2075=i ncome on the farm. 
.-. 5%=$2075, 

1%=4 of$2075=$415, and 
100$ =100 times $415=$41500=value of farm. 
$11500 — $500=$41000=amount invested in stock. 
100%=par value of the stock. 
102%=market value, or amount invested. 

-^%=brokernge 
102%+£%==102£%==entire cost of stock. 
... 102|%=$41000, 

1%=^ of $41000=$400, and 

1^2 [railroad stock. 

100%=100 times $100=$40000=par value of the 
1. $2075+$2025=$4100=incoroc on railroad stock. 
•2. $40000=100%. 

3. $l%=nHhnr of 100%= T io c /r , and [dend. 

4. $4100=4100 times -^%=10:j;%=rate of divi- 
-. 10J% =i ate of dividend. (R. H. A., p. 22^ prod. £.) 

What must be paid for 6% bonds to realize an income of 

S% ° n the investment? 
L. 100%=amount invested. 

I. 6%=income on the par value of the bonds. 
5. 8%=income on the investment, 
i. .-. 8% of investment=6% of the par value, 
). 1% of investment^ of 6%=f % of the par value, and 
). 100% of investment=100 times f%=75% of par value. 
•. Must pay 75% to make 8% on the investment. 
Note. — It must be borne in mind that 100% of any quantity is 
the quantity itself. .\ 100% of the amount invested equals the 



(9.) 



III. 
I. 



II. 



III. 



FINKEL'S SOLUTION BOOK. 



amount invested. It must also be remembered tbat the income 
o«i the par value is equal to the income on the investment. Sup- 
pose I buy a 500-dollar 6% bond for $400. The income on the 
par value, or face of the bond is 6% of $500, or $30. But $30 is 
lY/o of $400, the amount invested. Hence, the truth of step 4 
in the above solution. 



I. 



Which is the better investment, buying 9% stock at 
25% advance, or 6% stock at 25% discount. 



\\.{ 



ii. 



(10 

(20 
(3.) 

(40 
(5.) 

(60 
(70 



1 (8.) 



100%=amount invested in the 9% stock, 
100%=par value. 
25 %— premium. 
100%+25%=125%=market value. 
.-. 125%=100%, 

1%=^ of 100%=4%, and 
100%=1 00 times |%=80%=par value in terms 
of the investment. 
100%=80%, 

l%=rb of 80%=|%, and 
3. 9%=9 times |%=7-J-%=income 






[stock, 
of 9% 



B. 



(1.) 100%=amount invested in 6% stock. 
(2.) 100%=par valueof 6% stock. 
(3.) 25%=discount. 
(4.) 100%— 25%=75%=market value. 
(5.) ••• 75%=100%. 
(6.) 1%=A of 100%=li%, and 
(7.) 100%=100 times l£%=133£%=par value of 
the 6% stock in terms of the investment, 
rl. 100%=133£%, 
(8.W2. l%= T i Tr of 133i%=li%, and [stock. 

13. 6%=6 times l|%=:8%=income of 6% 
III. .-. The latter is the better investment, since it pays 8% — 
7^%, or \°Jo more income on the investment. 

( Greenleafs N. A., p. 298, f rob. 5.) 

I. If I pay 87^-% for railroad bonds that yield an annual in- 



come of 7%, what % do I get on my investment 



(10 

(2.) 

(30 
(40 

(5.) 
(60 



(V 



4 

13. 



III. 



100%=investment. 

100%=par value. 

87^%=market value, or amount invested. 

.-. 87|%=100%, from (1.) 

1%=^ of 100%= 1|%, and 

100%=100 times l|%=114|-%=par value in 

terms of the investment. 
100%=114f%, 

1%=^ of 114f %=li%, and [ment. 

7%=7 times l^%=8%=income on the invest- 
:income on the investment. 



STOCKS AND BONDS. 



75 



A banker owns 2|% stocks at 10% below par, and 3% 
stocks at 15% below par. The income from the former 
is 66f % more than from the latter, and the investment 
in the latter is $11400 less than in the former; required 
the whole investment and income. 



(3.) 



(5-) 



((>.) 



(7-) 



100%=investment in the former. 

100% — $11400=investment in the latter. 

100%=par value of the former. 
10%=discount of the former, [vested in former. 

100% — 10%=90%— market value, or amount in- 

.-. 90%=100%, from (1), 

1%=TU of 100%=H%, and 

100%=100 times 1-L%=llli%=par value of for- 
mer in terms of the investment. 

100%=llli%, 

1%=T*TT of 11H%=H%, and 
2|%=2£ times l^%=2J%=income of former in 
terms of the investment. 

100%=par value of the latter. 
15%=discount. [vested in the latter. 

100% — 15%=85%=market value, or amount in- 

.-. 85%=100%— $11400, from (2), 
1%= ¥ V of (100%— $11400 )=1 J- 

100%=100 times ( l T 3 y %_$134 T 2 7 ) =11711%- 
$13411^-f=par value of latter in terms of former. 

100%=117-H%— $13411jf, [$134 T V, and 

1% =t*t of (H7ff%-$134111f)= 1 T 3 T %- 
3%=3 times (l T y%— $134 1 V)=3/ T %— $402 T 6 T 
=income of latter in terms of the investment. 
1007^=income of the latter. 
100%+66f%=166f%=income of the former. 
2-/ f %=income of the former. 
.-. 166f %=2J%, 
1 



(11.) 



14. 



■■U\%- 



■$134 T V, 



1%: 



of 2$%=^%, and 



;i [terms of income of former. 

16. 100%?=100 times ^%=lf %^=income of latter in 
3 T 9 T %— $402 T 6 T =income oif the latter. 
... 3 T » T %-$402 T <V=lf %. 
144 %== $402 T « T %, 

1 %=pi of $402 T V=$216, 

100%=100 times $216=$21600= 
100%— $11400= $21600 — $11400 
vestment in latter. 
2^%=2^ times $216=$600=income of former. 



[former, 
investment in 
= $10200= in- 



76 



FINKEL'S SOLUTION BOOK. 



III. 
I. 



II. 



(15.) 3 T 9 T %— $402 T 6 T =3 T 9 T times $216— $402- 1 6 T =$360= 

income of latter. 
( 16. ) $21600-|-$10200=$31800=whole investment. 
(17.) $600-f-$360=$960= whole income. 

. f$31800=whole investment, and 

•|$960=whole income. (JR. H. A., p. 225, firob. £ ) 

W. F. Baird, through his broker, invested a certain sum 
of money in Philadelphia 6's at 115-J-%, and three times 
as much in Union Pacific 7's at 89-J%, brokerage \<f m 
both cases; how much was invested in each kind of stock 
if his annual income is $9920? 

(1.) 100%=amount invested in Philadelphia 6's. 

(2.) 300%=amount invested in Union Pacific 7's. 
1. 100%=par value of Philadelphia 6's. 



(3.)J 



115^%— market value. 
-^%=brokerage. 

115£%+|%=116%=entire cost of Phila. 6's. 
... 116%=100%. 
1%=^ of 100%=H%, and 

100%=100 timesff%=862 § 9%=par value of Phil- 
adelphia 6's in terms of investment. 



(4.). 



(5.) 



(6.)- 

(70 
(8.) 
(9.) 

(10.) 

(11.) 
(12.) 



1. 100%=86<&%, 

2- l%=droOf86A%,and 

3. 6%=6 times ||-^=5^%=income of Philadel- 
phia 6's in terms of investment. 

1. 100%=par value of Union Pacific 7's. 

2. 89|%=market value. 

3. -L%=brokerage. 

4. g9 i%_^L% == 90%=entire cost of Union Pacific 7's. 

5. .-. 90% =300%, 

6- !%=■& of300%-=3i%, and 

7. 100%=100 times 3|%=333i%=par value of 
Union Pacific 7's. 

1. 100%=333J%, 
2- l%=- 1 -io of 333i%=3i%,and 
3. 7%=7 times 3i%=23-J-%=income of Union Pa- 
cific 7's in terms of investment. 

5^%+23|-%==28f±%:=whole income. 

$9920=whole income. 

... 28|f%=$9920, 

l%=^of$9920=$348, 



[in Philadelphia 7's. 
100%=100 times $348=$34800=amount invested 
300%=300 times $348=$104500= amount in- 
vested in Union Pacific 7's, 



STOCKS AND BONDS. 



77 



III. •• 



$34800=amount invested in Philadelphia 6's, and 
$104400=amount invested in Union Pacific 7's. 

(R.H.A.^p. 225, fir ob. 6.) 



I. Thomas Reed bought 6% mining stock at 114^-%, and 
4% furnace stock at 112%, brokerage \ c / ; the latter 
cost him $430 more than the former, but yielded the 
same income ; what did each cost him? 



II. 



III. 



(1.) 

(2.) 



(3.) 



(4-.). 



100%=amount invested in mining stock. 
100%+$430=amount invested in furnace stock. 

1. 100%=par value of mining stock. 

2. 114^%=market value. 

3. £%=brokcrage. 

4. 114+%+i%=115%=entire cost. 

5. .'. 115%=100%, from (1), 

6. l%= T h of 100%=H%, and 

7. 100%=100 times f£%==96f|%==P ar value of 

mining stock in terms of investment. 
1. 100%=96ff%, 
2- l%=iiir of 96||%=|t%, and 
3. 6%=6 times fj%=5^%= 



:income of mining 



stock in terms ot investment. 

1. 100%=par value of furnace stock 

2. 112%=market value. 

3. ^-%=brokerage. 

4. 112%-H%=112|%=entire cost. 

5. .-. llg£%=100%+$430, 

l%== T 4r of ( 1 00%+$430) 



6. 



-4 r ' 



7o+^^ 



1 1 .) i ■ "XTi and 

100%=100 times (£%+$3H)=88|%+$382f== 
par value of furnace stock in terms of investm't. 

100%=88f%+$382J, 

l%=rJ-o of (88f%+$382J)=|%+*3H, and 

4%=4 times (f%+$3fJ)=3f%+$15i|=income 

of furnace stock in terms of the investment. 
.-. 5/3 % =3 J % +$1 5^f , by the conditions of the 

problem, 
H^=$15if, 

l#=TTvF of $15J*=$9.20, and _ . . f . 
I 1 ,;',; ° [mining stock. 

1 00 9f>=^-. 1.00 times $9.20=$920=amount invested in 

100% +$430==$1350==amount invested in furnace 

stock. (jR. H. A., p. 225,prob. 7.) 

, $920=amount invested in mining stock, and 
" *l$1350=amount invested in furnace stock. 



78 



FINKEL'S SOLUTION BOOK. 



I. 



II. 



III. 



Suppose 10% state stock is 20% better in market than 4% 
railroad stock; if A."s income be $500 fronreach, how 
much money has he paid for each, the whole investment 
bringing 6-g-f 3 % ? 



(10 



(20 



100%=par value of state stock. 

10%=income. 

$500=income. 

... 10% =$500, 
1%= T \ of $500=450 
100%=100 times $50 



and [stock. 

=$5000=par value of state 



100%=par value of railroad stock. 
4%=income. 
$500=income. 
... 4%=$500, 

l%=i of $500=4125, and [railroad stock. 

100%=100 times $125=$12500=par value of 
(3.) " $5000=| of $12500, i. e., the face of state stock 
is \ of face of railroad stock. 
100%=whole investment. 
6-g-|~^%=income of whole investment. 
$500+$500-=$1000=income of whole investment. 
■•■ 63l3%=$1000, 

l%= 7r 4- of $1000=$166.50, and 



(40 



6 



(5.) 



100%: 
100%: 
(1 



=100 times $166.50=$1G850= 
=investment in railroad stock 



[ment. 
whole invest- 



40%=f of 100%=inyestment in state stock, 

excluding the 20% excess. 
100% =40%, 

1%=t*o of 40%=|%, and 
20%=20 times |% = 8%=excess of state 
stock over same amount of railroad stock. 
40%+8%=48%==investment in state stock. 
100%+48%=148%=whole investment. 
$16650=whole investment. 
.-. 148%=$16650, 

l%= T ig of $16650=$112 50, [railroad stock. 

100%=100 times $112 50=$11250=investment in 
48%=48 times $112.50=$5400=investment in 
state stock 

{$11250=amount invested in railroad stock, and 
$5400=amount invested in state stock. 

(/?. H. A., p. 227,prob. 5.) 



9 ) 



EXAMPLES. 79 



EXAMPLES. 

1. What could I afford to pay for bonds yielding an annual 
income of 9% to invest my money so as to realize 6% on the in- 
vestment? Arts. 150%. 

2. What must I pay for Chicago, Burlington & Qiiincy Rail- 
road stock that bears 6% that my annual income on the invest- 
ment may yield 5% ? Ans. 120%. 

3. Bought 75 shares N. Y., P. & O. Railroad stock at 105%, 
and sold them at 108-^% ; how much did I gain in the transac- 
tion? Ans. $262.50. 

4. How many shares of bank stock at 5% premium, can be 
bought for $7665? Ans. 73. 

5. A broker bought stock at 4% discount, and, selling them 
at 3% premium, gained $1400 ; how many shares did he buy? 

Ans. 200. 

6. At what price must I buy 15% stock that it may yield the 
same income as 4% stock purchased at 90% ? Ans. 337-J%. 

7. Hjw much must I pay for New York 6's so that 1 may 
realize an income of 90%. Ans. 66f %. 

8. At what price must I buy 7% stock so that they may yield 
an income equivalent to 10% stocks at par? Ans. 70%. 

9. What sum must [ invest in U. S. 6's at 112% to secure an 
annual income of $1800? Ans. $35400. 

10. Which is the more profitable, and how much, to invest 
$5000 in 6% stock purchased at 75%, or 5% stock purchased at 
60%?? Ans, The latter; $16f. 

11. If a man who had $5000 U. S. 6's of 1881 should sell them 
at 115%, and invest in U. S. 10-40's purchased at 105%, would 
he gain or lose and how much? Ans. Loss $26.19. 

12. When gold is at 120, what is a "greenback" dollar worth ? 

Ans. 83i/. 

13. Suppose the market value of 5% bank stock to be 11t,-% 
higher than 8% corporation bonds ; I realize 8% on my invest- 
ment, and my income from each is $180; what did I invest in 
each? . 'Ans. $2923.07-^ in former, and $1576.92 T \ in latter. 

14. A bought 0% railroad stock at 109£%, and 4£% pike stock 
at 107|%, brokerage -J-% ; the former cost $100 less than the latter 
but yielded the same income; what did each cost him? 

Ans. $1100 cost of former, and $1200 cost of latter. 



80 FINKEL'S SOLUTION BOOK. 

15. What rate % of income shall I receive if I buy U. S. 5's at 
a premium of 10%, and receive payment at par in 15 years? 

Ans.Z\\o[ G . 

16. Suppose the market value of 6% corporation stock is 20% 
less than 5% state stock; if my income be $1200 from each, what 
did I pay for each if the whole investment brings 6% ? 

Ans. $16000, and $24000. 

17. I bought 2^% stock at 80%, and 4J% stock at 86%. 
The income on the former was 44f % more than on the latter, 
but my investment is $22140 less in the latter than in the former; 
what do I realize on my investment? Ans. 3||^%. 

Hint. — Find the whole investment, and whole income as in the problem 
on page 75. Then find what % the whole income is of the whole invest- 
ment. 

18. Invested in U. S. 4|-'s at 105, brokerage J% ; f as much 
in U. P. 6's at 119f, brokerage ^%; and 3 times as much in N. 
Y. 7's, brokerage ^%. If my entire income is $1702, find my 
investment. Ans. $25320. 

19. A. paid $1075 for U. S. 5-20 6% bonds at 7|% premium, 
interest payable semi-annually in gold. When the average pre- 
mium on gold was 112%, did he make more or less than B. who 
invested an equal sum in railroad stock at 14% below par, which 
paid a semi-annual dividend of 4% ? 

Ans. A. makes $16.40 less than B. every six months. 

20. I invested $4200 in railroad stock at 105, and sold it at 
^j%; how much must I borrow at 4% so that by investing all I 
have in 6% bonds at 7% interest, payable annually, I may re- 
trieve my loss in one year? Ans. $18600. 



VII. INSURANCE. . 

1. Insurance is indemnity against loss or damage. 

I-, -d , T M. Fire Insurance. 

1. Property Insurance. \ n ,,»- . T 

r J (2. Marine Insurance. 

( 1. Life Insurance. 

2. Personal Insurance. < 2. Accident Insurance. 

( 3. Health Insurance. 

3. Property Insurance is the indemnity against loss or 
damage of property. 

4. Personal Insurance is indemnity against loss of life 
or health. 

5. Fire Insurance is indemnity against loss by fire. 



INSURANCE. 81 

6. Marine Insurance is indemnity against the dangers 
of navigation 

7. Life Insurance is a contract in which a company 
agrees, in consideration of certain premiums received, to pay a 
certain sum to the heirs or assigns of the insured at his death, or 
to himself if he attains a certain age. 

8. Accident Instirance is indemnity against loss by 
accident. 

9. Health Insurance is a weekly indemnity in case of 
sickness. 

10. The Insurer, or Underwriter, is the party, or 
company, that undertakes the risk. 

11. The Risfc is the particular danger against which the 
insurer undertakes. 

12. The Insured is the party protected against loss. 

13. The Premium is the sum paid for insurance; and is 
a certain per cent, of the amount insured. 

14. The Amount, or Valuation, is the sum for which 
the premium is paid. 



I. My house is permanently insured for $1800, by a deposit 
of ten annual premiums, the rate per year being |%; 
how much did I deposit, and if, on terminating the in- 
surance, 1 receive my deposit less 5% ; how much do I 
get? 

(1.) 100%=$1800, 

(2.) l%= T io of $1800=$18, and 

(3.) !%=j times $18=$13.50=ane annual deposit. 

(4.) $135=10 times $18.50=ten annual deposits. 

( 1. 100%=$135, 
(5.W2. l#=rfo of $135=$1.35, and 

(.3. 5% =5 times $1.35=$6.75=deduction. 
(6.) $135— $6.75=$128.25=what I received. 



II. 



TTT i $135=amount deposited, and 
' ' ' } $1 28. 25=a mount received. 

(7?. H. A., p. 230, p rob. 5.) 

I. An insurance company having a risk of $25000, at t 9 q%, 
reinsured $10000, at i% , with another office, and $5000, 
at 1%, with another; how much did it clear above what 
it paid? 



II. 



iu 



82 FINKEL'S SOLUTION BOOK. 

(1.) 100%=$25000, 
(2.) l a /c=-rio of $25000=$250, and 

(3.) T \(f = T \ times $250=$225=what the company 
received for taking the risk. 
'1. $10000=amount the company reinsured at i%. 
2. 100%=$10000, 
(4.W3. l%= T io of $10000=$100, and 

4. f %=f times $100=$80=what the company paid 
for reinsuring $10000. 

1. $5000=amount reinsured in- another office at 1%. 

2. 100% =$5000, [for reinsuring $5000. 
(5.)<[3. lf =ifar of $5000=$50=what the company paid 

4. $S0+$50=$130=what the company paid out. 
15. $225— $130=$95=what it cleared. 

III. .*. $95=what the company cleared. 

(R. H. A., p. 230, prod. 7.) 

I. I took a risk at 4-|-% ; reinsured -| of it at 2%, and ^ of it 
at 2-|%; what rate of insurance do I get on what is left? 

(1.) 100%=wholerisk. 
(2.) l-£-%=premium. 

rl. 40% =4 of 100%=amount reinsured at 2%. 
,„ J 2. 100%=40%, 
1 ; |3. l%= T i^of40%=|%, and [suringfof the risk. 

(.4. 2%=2 times -|%=|%=anaount I pay outforrein- 

(1. 25%=£ of 100%=second part reinsured. 

I 2. 100% =25%. 
(4.)(3. l%= T h of 25%=i%, and 

4. 2£%=2£ times i%=|%=amount I paid out for 

^ reinsuring ^ of the risk. 

(5.) !%-J-!%=li-2-%=amount of premiums paid out. 

(6.) H%— H-J%=^ r ^=amount of premium I had left. 
(7.) 40%-f-25%=65%=whole amount reinsured. 

(8.) 100%— 65%=35%?=risk left on which I received 
To% premium. 

rl. 35%=100% of itself. 
(9.)<J2. l%=-^of 100%=2f%,and 

^3. ■$r.% =p & ti mes 2f %= T 3 ¥ %=rate of premium 
III. .*. T 3 :f %=rate of insurance I receive. 

(R. H. A., p. 231, prod. 6.) 

Remark. — 35% is the base and -%§°fo is the percentage, and we 
wish to know what per cent. -}^°/o is of 35%. 

I. Took a risk at 2%; reinsured $10000 of it at 2^% and 
$8000 at lj%; my share of the premium was $207-50 ; 
what sum was insured? 



INSURANCE. 



S3 



II. i 



(1.) 



A. 100%=$10 

h. i%= r i T oi 

U. 24%=2^tii 



100%=$10000, 

,f $1000G=$1(X>, and [$10000 reinsured. 

8%=2-i times $100=$212.50=amount paid out on 

rl. 100%=$8000, 

(2.)<[ 2. l°/o=jh of $8000=$80, and [$8000 reinsured. 

13. If %=lf times $80=$140=amount paid out on 



(3.) $212.504-$140=$352.50= 

(4.) $207.50=what I realize. 

(5.) .-. $352.50+207.50=$560: 

(6.) 100%=risk. 

(7.) 2%=premium. 

(8.) $560=premium. 

(9.) .-. 2%=$560, 

(10.) l%=i of $560=$280, and 

(11.) 100^=100 times $280=$28000: 



whole amount paid out. 
=premium on whole risk. 



-risk. 



III. 



$28000=risk. 



(/?. II A., p. 232,prob. 6.) 



I. I can insure my house for $2500 at -f^fy premium annually, 
or permanently by paying down 12 annual premiums; 
which should I prefer, and how much will I gain by it 
if money is worth 6% per annum to me? 



II 



(1.) 

(2.) 
(3.) 

(4.) 



(5.) 

((>.) 
^ (7.) 



100%=$2500. 

1%=^ of $2500=$25, and 

^c? =z-^ times $25=$20=one annual premium. 

$240=12 times $20=twelve annual premiums. 

1. 100%=the amount that will produce $20 an- 

nually at 6%. 

2. 6%— interest. 

3. $20=interest. 

4. .*. 6%=$20, 

5. 1%=£ of $20=$3^, and 

6. 100%— 100 times $8£=$333£=the amount that 

will produce $20 annually at 6%. 
$333£+$20=$3534=amount I would have to pay 

down by the former condition. [tion. 

.-. $353j — $240=$113|=gain by the latter condi- 



III 



(The latter is the better. 
( $113£=gain. 



Remark. — In (6) we add $20, since a payment must be made 
immediately. $33&J will not produce that sum until the end of 
the year. 



84 FINKEL'S SOLUTION BOOK. 

I. The Mutual Fire Insurance Company insured a building 
and its stock for f of its value, charging If °f c . The 
Union Insurance Company relieved them of \ of the 
risk, at 1^%. The building and stock being destroyed 
by fire, the Union lost $49000 less than the Mutual; 
what amount of money did the owners of the building 
and stock lose? 

100%=value of the building and stock. 
66f %=f of 100%=amount insured. 
lf%=rate of insurance. 
10.0%=66f%, 

1 %=dnr of 66f %=f % , and 

lf%=lf times f%=li%=what Mutual received 
from the owners of the building and stock. 
(5.) 16f %=i of 66f%=amount of which the Union 
relieved the Mutual, 
fl. 100%=16f%, 
(c J 2. 1%= T U of 16f%=i%, and 
v ; |3. 1|%=1| times .*.%=£%=what the Mutual paid 

L the Union for taking the risk of 16f %. 
(7.) 16f%+li%==17f%=whole amount the Mutual 
received. [paid out. 

(8.) 66f%+i%=66H%=whole amount the Mutual 
(9.) .*. 66H%— 17f%= 49^2% = amount the Mutual 
lost. 
II.<! (10-) 16|%=amount the Union paid the Mutual. 

(11.) ^%=amount the Union received from the Mutual. 
(12.) .'. 16f %— i%=16 T 5 2%=amount the Union lost. 
(13.) 49 T V%— 16 T 5 2%=32f%=what the Mutual lost 
more than the Union. [Union. 

(14.) $49000=what the Mutual lost more than the 
(15.) .'. 32f %=$49000, 

(16.) 1%=^ of $49000=11500, and _. 

v J ' 32f [ing and stock. 

(17.) 100%=100 times $1500=$ 150000= value of build- 
(18.) 66|%=66f times $1500=$100000=amount in- 
sured, [ers lost, it not being insured. 
33^%=33i times $1500=$50000=what the own- 
100%=$100000, 

1%=^ of $100000=$1000, and 
If %=lf times $1000=$1750=what the owners 

paid the Mutual for insurance. 
.-. $50000 + $1750= $51750=whole amount the 
owners lost. 
III. .-. The owners of the building and stock lost $51750. 




EXAMPLES. 85 



EXAMPLES. 

1. At lf%, the premium for insuring my store was $89.10; 
what was the amount of the insurance? ■ Ans. $6480. 

2. The premium for insuring a tannery for f of its value, at 
!■§■%> was $145.60; what was the value of the tannery? 

Ans. $11648. 

3. A store and its goods are worth $6370. What sum must 
be insured, at 2%, to cover both property and premium? 

Ans. 

4. The premium for insuring $9870 was $690.90 ; what was 
the rate? Ans. 7%. 

5. A merchant whose stock of goods was valued at $30000, 
insured it for f ot its value, at f %. In a fire he saved $5000 of 
the goods. What was his loss? What was the loss of the in- 
surance companies? Ans. 

6. A man paid $180 for insuring his saw mill for |- 'of its 
value at 3%; what was the value of the mill? Ans. . 

7. A house which has been insured for $3500 for 10 years, at 
■|% a year, was destroyed by fire ; how much did the money re- 
ceived from the company exceed the cost of premiums? 

Ans. . 

8. Took a risk on a house worth $40000, at 2%; reinsured 
-J of it for 2i%, and ^ of it at 2^% ; in each case the amount cov- 
ers premium; how much do I gain? Ans. $99,558. 

9. Took a risk at lj%; reinsured f of it at \\ c /o \ my share 
of the premium was $43 ; what was the amount of the risk? 

Ans. $17200. 

10. Took a risk at 2£% 5 reinsured \ of :'t at a rate equal to 
3% of the whole, by which I lost $37.50. What was the value 
of the risk? Ans, $5000. 



86 FINKEL'S SOLUTION BOOK. 

CHAPTER XIII. 

INTEREST. 

' I. SIMPLE INTEREST. 

1. Interest is money paid by the borrower to the lender 
for the use of money. 

%, The Principal is the sum of money for which interest 
is paid. 

3. The Mate of interest is the rate per cent, on $1 for a 
certain time. 

4. The Time is the period during which the money is on 
interest. 

5. The Amount is the sum of the principal and interest. 

6. Simple Interest is interest on the principal only. 

7. Legal Interest is at the rate fixed by law. 

8.- Usury is interest at a rate greater than that allowed by 
law. 

Let P—the principal, 

r=the interest on $1 for one year, 
jR=l-\-r=ZL mount of $1 for one year, 
n=the number of years, 
^4=amount of P for n years, 
/V=simple interest on P for a year, 
Pnr=simple interest on P for n years. 
' P-\-Pnr=P (l-f-^r)=amountof P ioxn years. 
^4=amount of P for n years. 
/, A=P+Pnr=P(l+mr) (I.); 

.'. P=TT- ....(II.); 
... Pnr=A—P. 

nr nr x 

.: r=±£ .... (IV.); and 

. ••• «=^ ■■■■ (v.)- 

When any three of the quantities A, P, n, r are given, the 
fourth may be found. 

CASE I. 

{Principal,^ 
Rate, and >to find the interest. Formula, I=Prn. 
Time, J 



SIMPLE INTEREST. 87 

I. Find the interest of $300 for two years at 6%. 
By formula, 

Interest /V/z=$300X-06x2=$36. 
By 100% method. 
rl. 100% =$300, 
n I 2. 1%= T U of $300=$3, and 
±J -i3. 6%=6 times $3=$18=interest for one year. 

U. $36=2 times $18=interest for 2 years. 
III. .-. $36=interest on $300 at 6% for 2 years. 

CASE II. 

rPrincipal,>| A—P 

Given<Rate, and >to find the time. Formula, n- 



Pr 

^Interest, J 

I. In what time, at 5%, will $60 amount to $72? 
By formula, 

A—P $72— $60 , 
te -pT = $60xT05 =4yearS - 
By 100% method. 
'1. $72=amount. 

2. $60=principal. 

3. $72— $60=$12=interest for a certain time. 
II.H 100% =$60, 

5. l%= T i ? of$60=$f, and 

6. 5%=5 times $i=$8=interest for one year. 

7. $12=interest for 12--3, or 4 years. 

III. V. $60 at 5% will amount to $72 in 4 years. 

CASE III. 

rPrincipal,^ A—P 

Given^Time, and >to find the rate. Formula, r= — - — . 

llnterest, J Pn 

I. I borrowed $600 for two years and paid $48 interest; what 
rate did I pay? 
By formula, 

A—P I $48 A . . . 

^ pn -pn =moxr =M== ^ ' 

By 100% method. 

1. $48=interest for 2 years. 

2. $24=| of $48=interest for 1 year. 
II.{3. $600=100%, 

4. $l= ¥ i ir ofl00%=i%,and 

5. $24=24 times i%=4%. 
III. .'. I paid 4% interest. 



FINKEL'S SOLUTION BOOK. 



CASE IV. 

h 1 

Given<[Rate, and >to find the principal. A — P 

est. J 



{Time, 
Rate, 
Interest. J Formula, P=~ 



I. The interest for 3 years, at 9%, is $21.60; what is the 
principal? 

By formula, 

p _ A—P _ 7_$21.60 



II. 



nr nr 3X.09 
By 100% method. 

4. $21.60=interest for 3 years. 

2. $7.20=4 of $21.60=interest for 1 year. 

3. 100%=principal. 

4. 9%=interest for 1 year. 

5. $7.20=interest for 1 year. 

6. .-. 9%=$7.20, 

7. 1%=4 of $7.20=$.80, and 
^8. 100%=100 times $.80=$80=principal. 

III. .-. $80=the principal. 

CASE V. 

/Time, v A 

Given<Rate, and>to find the principal. Formula, />= — = .. 

lAmount J l + nr 

I. What principal will amount to $936 in 5 years, at 6%? 

By formula, 

A $936 r =$720. 



l+» r 1+5 X -06 
By 100% method. 

4. 100%=principal. 

2. 6%=interest for 1 year. 

3. 30%=5 times 6%=interest for 5 years. 

4. 100%+30%=130%=amount. 

5. $936=amount. 

6. .-. 130%=$936, 

7. l%= T i 7 of$936=$7.20, and 

8. 100%=100 times $7.20=$720=principal. 

III. .*. $720=the principal that will amount to $936 in 5 years 
at 6%. 



IU 



TRUE DISCOUNT. 



I. In what time will any sum quadruple itself at 8% ? 

II.< 

III. .'. Any principal will quadruple itself in 37-J- years at 8%. 



fl. 100%=principal. Then 

2. 400%=the amount. 

3. ... 400%— 100%=300%=interest. 

4. 8%=interest for 1 year. 

5. 300%=interest for 300-J-8, or 37| years. 



II. TRUE DISCOUNT. 

1. Discount on a debt payable by agreement at some 
future time, is a deduction made for "cash," or present payment; 
and arises from the consideration of the present worth of the debt. 

2. Present JVovtJl is that sum of money which, put on 
interest for the given time and rate, will amount to the debt at 
its maturity. 

3. True Discount is the difference between the present 
worth and the whole debt. 

Since P will amount to A in n years, P may be considered 
equivalent to A due at the end of n years. 

.-. P may be regarded as the present worth of a given future 
sum A. 

••• p =^r~ 

l-\-nr 

I. Find the present worth of $590, due in 3 years, the rate 
of interest being 6%. 



By formula, 

A $590 



;=$500. 



l+nr 1+3 X- 06 
By 100% method. 

1. 100%=present worth. 

2. 6%=interest on present worth for 1 year. 

3. 18%=3x6%=interest for 3 years. 

4. 100%+18%=118%=amount, or debt. 

5. $590=debt. 

6. .-. 118%=$590, 

7. 1%=^ of $590=$5, and 

8. 100%=100 times $5=$500=present worth. 

III. .*. $500=present worth of $590 due in 3 years at 



II. 



90 



FINKEL'S SOLUTION BOOK. 



A merchant buys a bill of goods amounting to $2480; he 
can have 4 months credit, or 5% off for cash : 'if money 
is worth only 10% to him, what will he gain by paying 
cash? 



II. 



(1.) 100%=present worth of the debt. 

(2.) 10%=interest on present worth for 1 year. 

(3.). 3^%=interest for 4 months. 

(4.) 100%+3J%=103£%=amount of present worth, 

which equals the debt, by definition. 

(5.) $2480=the debt. 

(6.) .-. 103i%=$2480, 

(7.) 1 %=j^t of $2480=$24, and 

(8.) 100%=100 times $24=$2400=present worth. 
(9.) $2480— $2400— $80=true discount, 
rl. 100%=$2480. 
(10. K 2. 1%= T ^ of $2480=$24.80, [count for cash. 

13. 5%=b times $24.80=$124=trade discount, or dis- 
(11.) .'. $124— $80=$44=his gain by paying cash. 



III. .*. He would gain $44 by paying cash. 

(J?. 3d p., p. 258, prod. 10.) 



Remark.— -It is clear that $2480— $124,=$2356 would pay for 
the goods cash. If the merchant had this sum of money on hand, 
it would, in 4 months, at 10%, produce $78. 53-J- interest. But if 
he pays his debt he will make $124. Hence he will gain $124 — 
$78.53i=$45.46|. 



III. BANK DISCOUNT. 

1. Hafih Discount is simple interest on the face of a 
note, calculated from the day of discount to the day of maturity, 
and paid in advance. 

2. The Proceeds of a note is the amount which remains 
after deducting the discount from the face. ? 



Given 



Face of note, 
Rate, and 
Time, 



CASE I. 

to find the discount and proceeds. 

Formula * ^=^X-> 



-\ 



P=F—D. 



BANK DISCOUNT. 91 

I. What is the bank discount of $770 for 90 days, at 6% ? 
By formula, 

D=Fx r X ^=$770 X .06 X (9 °+ 8) = $l 1.935. 

By 100% method. 
fl. 100%=$770, 
Tr 1 2. 1%= T U of $770=$7.70, and 
iL 'l3. 6%=6 times $7.70=$46.20=discount for 1 year. 

U. $11.935=3%V of $46.20=discount for 93 days. 
III. .-. $11.935=bank discount on $770 for 90 days at 6%. 

CASE II. 

C Proceeds, ) 

Given < Time, and > to find the face of the note. 

(Rate, ) P 

rormula, jt=> 



1 — rn 
I. For what sum must a note be made, so that when dis- 
counted at a bank, for 90 days, at 6% the proceeds will 
be $393.80? 
Bv formula, 

P _ $393.80 

By 100% method. 

1. 100%=face of the note. 

2. 6%=discount for one year. 

3. lH%=inftF of 6%=discount for 93 days. 

4. 100% — Hi%=982 e o%=proceeds. 
II.<[5. $393.80=proceeds. 

6. /. 98^%=$393.80, 

7. 1%=^ of $393.80=$4, and 

K S. 100%=100 times $4=$400=face of the note. 
III. .'. $400=face of the note. . 

CASE III. 

Given rate of bank discount, to find the corresponding rate of 
interest. Formula, rate of /.=- 



r 



1 — rn 

I. What is the rate of interest when a 60 day note is dis- 
counted at 8% per annum? 
By formula, 

»* of / =T^= (l-^x.08) = - 08 ^ =8 ^^- 



&> FINKEL'S SOLUTION BOOK. 



By 100% method. 

1. 100%=face of note. 

2. 8%=discount for 1 year. 

3. 11%=^%% of 8%=discount for 63 days. 

4. 100%— lf%=98f%=proceeds. 

5. 98f%=100% of itself. 
1 



HJ 



6. l%=gp of 100%=fff%, and - 



8%=8 times so o ^ == 8^^ ==ra te of interest. 
III. .-. The rate of interest on a 60 day note discounted at 8% 
per annum=8 ¥ 5 ¥ 6 -3%. 

CASE IV. 

Given the rate of interest, to find the corresponding rate of 

-r, , rate of/. 

discount. formula, r=-—. -— 

l-j-^Xrateof /. 

I. What is the rate of discount on a 60 day note which yields 

10% interest? 

By formula, 

rate0f/ - - - 10 - 093 3 7_Q3 3 7 <* 

r — 1+^xrate of /" l+^_ x ^ — U*nrr— *nn7<>- 
By 100% method. 

1. 100%=proceeds. 

2. 10%=interest on proceeds for 1 year. 

3. 1-|%— ¥ 6_ 3 -g- of 10%=interest on proceeds for 63 days. 

4. 100%+lf %=101f %=face of note. 

5. 101|%=100% of itself. 
1 



II. 



1% =IoTf of 100 % == ^%' and 



7. 10%=10 times jo o % =9 ff \ %. 
III. " .-. The rate of discount=9ff^%. 
Note. — It must be borne in mind that the interest on the pro- 
ceeds is equal to the discounton the face of the note. 



IV. ANNUAL INTEREST. 

1. A.n7iual Interest is the simple interest of the princi- 
pal and each year's interest from the time of its accruing until 
settlement. 

I. No interest having been paid, find the amount due Sept. 
7, 1877, on a note of $500, dated June 1, 1875, with 
interest at 6%, payable semi-annually. 



ANNUAL INTEREST. 



2yr. 3 mon. 6 d. 



1 yr. 



1 yr. 3 mon. 6 d. 



9 mon .8 6.. 



3 m. 6 d 



1st 



2d 



a d 



II. 



f(l.) 100%=$500, 
(2.) l%=^h of $500=$5, and 



1877—9—7 
1875—6—1 



III. 



(3.) 6%=6 times $5=$30=simple interest 2—3—6 

for 1 year. 
(4.) ,$68=2 T 4 T times 930=simple interest for 2 years, 3 

months, 6 clays. 
(5.) $15=| of $30=semi-annual interest. 
fl. 100%=$15, 

2. 1%=^ of $15=$.15, and 

3. 6%=6 times $.15=$.90=interest on one semi- 
annual interest for 1 year. 

4. $3.885=4 'j-J times $.90=interest on one semi-annual 
interest for the sum of the periods each draws int. 

(7.) .'. $500+$68+$3.885=$571.SS5=amount of the note. 
,'. $571.885=amount of the note. 



(6.) 



Explanation. — At the end of six months there is $15 interest 
due ; and, since it \v;is not paid at that time, it drew interest 
from that time to the time of settlement, which is 1 yr. 9 mon. 
6 da. At the end of the next six months, or at the end of the 
first year, there is another $15 due; and, since it was not paid at 
that time, it drew interest from that time to the time of settle- 
ment, which is 1 yr. 3 mon. 6 da. In like manner, the third 
semi-annual interest drew interest for 9 mon. 6 da., and the 
fourth for 3 mon. 6 da. This is the same as one semi-annual 
interest drawing interest for the sum of 1 yr. 9 mon. 6 da., 
1 yr. 3 mon. 6 da., 9 mon. da., 3 mon. 6 da. In the dia- 
gram, the line A B represents 2 yr. 3 mon. 6 day., A 1 repre- 
sents the first year the note run, and 1-2 represents the second 
year the note run. Between A and 1 is a small mark that de- 
notes the semi-annual period ; also one between 1 and 2. By 
such diagrams, the time for which to compute interest on the 
simple interest may be easily found. 

I. The interest of U. S. 4 r /r bonds is payable quarterly in 
gold; granting that the income from them might be 
immediately invested, at 6%, what would the income 
on 20 1000-dollar bonds amount to in 5 years, with gold 
at 105? 



94 



FINKEL'S SOLUTION BOOK. 



5 yr. 


























$•200 for 4 yr. 9 mon. at 6$, 


; $200 for 4 yr. 6 mon. at 6#. 


j $200 for 4 yr. 3 mon. at 696- 


4 i 1 i 

Istl 2d| 3d I 4th 


5| 61 7 8 


&C. 

9I 10I 


.J 


12 


n\ 


14 


»l 


16 


n| 


18| 


19 | 


B 

20 


1 
















4 






5 



II. 



(2.) 
(3.) 
(4.) 
(5.) 
(6.) 
(7-) 



(8.) 



(9.) 
(10.) 

l(H-) 



$1000=par value of one bond. 

$20000=par value of 20 bonds. 

100%=$20000, 

1%= T ^ of $20000=$200, and 

4%=4 times $200=$800=income for one year. 

$4000=5 times $800=income for five years. 

$200=i of $800=interest due at the end of first 
quarter, and which draws interest to time of set- 
tlement. 

n. ioo%=$2oo, 

2. 1%= T ^ of $200=$2, and [est for one year. 

3. 6%=6 times $2=$12=interest on quarterly inter - 

4. $570=47-^ times $12=interest on quarterly inter- 

est for the sum of 4f yr.+4-J- yr.-j-4£ yr.+ 

-f~i yr. , or 47^ years. 
.-. $4000+$570=$4570=income of bonds in gold. 
$1.00 in gold=$1.05 in currency. [rency. 

$4570 in gold=4570 times $1.05=$4798.50 in cur- 



III. .-. The bonds yield $4798.50 in currency. 

Explanation. — It must be borne in mind that the quarterly in- 
terest, $200, is put on interest at 6% as soon as it is due. At the 
end of the first quarter there is $200 due which draws interest at 
6% for the remaining time, 4 years, 9 months. The second 
quarterly interest is due at the end of six months and draws in- 
terest for the remaining time, 4 years 3 months, and so on with 
the remaining quarterly payments. This is the same as one 
quarterly payment drawing interest for the sum of 4f yr.-f-4-j- yr. 
+4i yr.-|-etc., or 47-J- years. 



I. What was due on a note of $1200, dated January 16, 1882, 
and due July 1, 1892, and bearing interest at 8%, 
payable annually, if the 2, 3, 5, and 7th years' interest 
were paid ? 



ANNUAL INTEREST. 



95 





fc96 for 9 yr. G m. 15 d. at 8#. 






>96 for 6 yr. 6 m. 15 d. 




J96, 4yr. 6 m. 15 d. 


i 




$96, 2 yr. 6 m. 15 d. 


i i 


$96,1 yr.6 m. 15 d. 




) 


) 




3 







$96,6 
1 


m 
5 rt. 



II. 



(2.) 
(3.) 
(4.) 
(5.) 
(6.) 
(7.) 



(8.) 



(9.) 



100% =$1200. 

l%= T £o of $1200=$12, and 

8%=8 times $12=$96=simple int. for one year. 

$480=5 x$96=five simple interests. 

$48=^ of $96=interest for 6 months. 

U= T \ of $48=interest for 15 days. 

.". $532=simple interest unpaid. 

100%=$96. 

1%= T ^ ¥ of $96=$.96, and [simple interest. 

8%=8 limes $.96— $7.68=intcrest on one year's 

$193.92=25£ times $7.68==interest on year's sim- 
ple interest for 9 yr. 6 mon. 15 da.,-|-6 yr. 6 mon. 
15 da. ,-(-4 yr. 6 mon. 15 da.,+2yr. 6 mon. 15 da., 
+1 yr.6 mon. 15 da. ,+6 mon. 15 da., or 25 yr. 
3 mon. 

.-. $532+$193.92=$725.92=amount of interest 

due. [1, 1892. 

$1200+725.92 = $1925.92 = amount due July 
$1925.92=whole amount due July 1, 1S92. 

V. COMPOUND INTEREST. 

1. Compound Interest is interest on a principal formed 
by adding interest to a former principal. 

Let 7 > =principal on compound interest. 
r=rate, 

7?=(l-)-;-)=a mount of one dollar for 1 year. 
p ( l-j-r)=/ ? i?=amount of P dollars for 1 year. 
P (l-j-r)' 2 = /Vi > '-=amount of P dollars for 2 years. 
P (l-\-r)* = PP"'— amount of P dollars for 3 years. 
P (l-j-r) n =/ > /? n ==amount of P dollars for n years. 
Let yl=amount of P dollars in n years, and 

/=the compound interest of P dollars for n years. 

Then I=PR n —P I. 

A=PP» II. 

■••-4 - 



(.(10) 



[. 



r; 



96 FINKEL'S SOLUTION BOOK. 

n _ A 

.'. R=yJ a_ IV. Applying logarithms to i? n =— , 

p P 

n log. 7?=log. A — log. P, whence 
log. A— log. P 

log. 7? ••' ' 
When compound interest is payable semi-annually. 
P (l-(-^-)=amount of P dollars for \ year. 
i 3 (1-j-f ) 2 =amount of P dollars for 1 year. 
P (1-j-f ) 2n — amount of P dollars for n years. 
.*. A—P (l-|-|-) 2n , when payable semi-annually. 
When compound interest is payable quarterly, 
P (l-|-^)=amount of P dollars for \ year. 
P (1+i ) 2 =amount of P dollars for \ year. 
P (l+i) 3 =amount of P dollars for f year. 
P (l-j-|) 4 =amount of P dollars for 1 year. 
P (1-j-f ) 4n =amount of P dollars for n years. 
.-. A=P (1+1)4-. 
When the interest is payable monthly, 

a=p (i+ T ' Y y*°. 

When the interest is payable q times a year, 
' A = P (l+^) qn - 

CASE I. 

C Principal, ) 
Given < Rate, and > to find the compound interest and amount. 

( Time, ) w ' ' ■■ i- i I=PP n —P, 

* Formulae, A==PjR n 

I. Find the compound interest and amount of $500 for 3 
years at 6%. 
By formulas, 

^==/>i? n =$500X(l+.06) 3 ==$595.508, and 
I=PR*— 7>=$500 x ( 1+.06 ) 3 — $500=$95.508. 
Remark. — In compound interest, the 100% method becomes 
very tedious. 

By 100% method. 
' (1.) 100%=$500, 
(2.) l%= T ^ r of$500=$5, 

(3.) 6%=6 times $5=$30=interest for 1 year. 
(4.) $500+$30=$530=amount, or principal for the 
second year. 

1. 100%=$530, 

2. l%=xi-o of $530=$5.30, [year. 
(5.)^ 3. 6%=6 times $5.30=$31.80=interest for second 

4. $530+$31.80=$561.80=amount, or principal for 
the third year. 



II. 



COMPOUND INTEREST. 97 

rl. 100%=$561.80, 

2. 1 %= T £o of $561.80=15.618, and [year. 

(6.)<J3. 6%=6 times $5.618=$33.708=interest for third 
4. 561.80+$33.708=$595.508=amount at end of the 
third year. 
(7.) $595.508— $500=$95.508= compound interest. 

t ( $95.508=compound interest, and 
I $595.508=compound amount. 



CASE II. 

Principal, 
Given <? Rate, and V to find the time. 



Compound Interest, ) _ , log. A — log./' 
1 ' Formula, n= to , -° . 

log. R 

I. In what time will $8000 amount to $12000, at 6% com- 
pound interest? 
By formula, 

log. ^4— log. P ^ Jog. 12000— log. 8000 
log. R log. 1.06 

4.079181—3.903090 a .. 1K , _.. , ■ 
=0 yr. 11 mon. Id da. We may solve the 

problem thus: $8000(1.06) n =$12000, whence (1.06) n =12000-f- 
8000=1.50. Referring to a table of compound amounts and 
passing down the column of 6%, we find this amount between 6 
years and 7 years. 

The amount for 6 years is 1.4185191 ; the amount for required 
time is 1.50. v. There is a difference of 1.50 — 1.4185191, or 
.0814809. The difference for the year between 6 and 7 is 
.0851112. .085lil2=amount for the whole period between 6 
and 7, .0814809=amount for %\\\\l of the period or, 11 mon. 15 
da. .-. The whole time=6 yr. 11 mon. 15 da. 



CASE III. 

C Principal, ^ 

Given < Compound Intesest or Amount, and V to find the rate. 
( lime, \ 



Formula, r=n \~ 1. 



I. At what rate, by compound interest, will $1000 amount 
to $1593.85 in 8 years? 

By formula^ _____ fr: V\j l ^r^ 




FINKEL'S SOLUTION BOOK. 



CASE IV. 



( Compound Interest or Amount ) 
Given < Time, and \ to find the principal. 



Rate. 



Formulae, P= 



A 
,or 

I 



R a —\ 

I. What principal, at compound interest will amount to 
27062.85 in 7 years at 4% ? 
By formula, 



CHAPTER XIV. 

ANNUITIES. 



I. An Annuity is a sum of money payable at yearly, or 
other regular intervals. 

II. Perpetual, or 
I'. CertSor 
4. Contingent. 

3. A Perpetual Annuity is one that continues forever. 

4. A Limited Annuity ceases at a certain time. 

5. A Certain Annuity begins and ends at fixed tinles. 

6. A Contingent Annuity begins or ends with the 
happening of a contingent event. 

7. An Immediate Annuity is one that begins at once. 

8. A Deferred Annuity is one that does not begin im- 
mediately. 

9. The Final or Forborne value of an annuity is the 
amount of the whole accumulated debt and interest, at the time 
the annuity ceases. 

•10. The Present Value of an annuity is that sum, 
which, put at interest for the given time and given rate, will 
amount to the initial value. 

II. The Initial Value of an annuity is the value of a 
deferred annuity at the time it commences. 



ANNUITIES. 



99 



CASE I. 

C Annuity, ) 
Given < Time, and > to find the initial value of a perpetuity. 

( Rate, ) 

What is the initial value of a perpetual annuity of $300 a 

year, allowing interest at 6% ? 
. 100%=initial value. 

6%=interest for 1 year. 

$300=interest for 1 year. 

.-. 6% =-$300. 

1%=4 of $300=$50, and 

100% = 100 times $50=$5000=initial value. 
Initial value=$5000. (R. H. A., p. 310, prod. 1.) 



I. 

IT 

III. 
I. 



fl 

2 
3 
-4 
5 
6 



What is the initial value of a perpetual leasehold of $2500 
a year payable quarterly, interest payable semi-annually 
at 6%; 6% payable annually ; 6% payable quarterly? 

1. Let .5== the annuity. Then .S=the amount due in 
3 months. 

2. ^S— [— -S^ l-)-^)=amount due in 6 months. 

3. /. ^ = S+S(l-r-.01i)==$625 + $625(1.0H) = 
$12r)9.37Y= amoun t due at the end of 6 months. 

4. 100%=initial value. 

5. 3%=semi-annual annuity. 

6. $1259.37^=semi-annual annuity. 

7. .-. 3%=*1259.37£. 

8. \%=\ of $1259.37i=$419.7916|, and 



A.<^ 



II.. 



B. 



19. 100%=100 times $419.79 16~initial value. 

1. Let 5"=amount due in 3 months. Then 

2. 5'-(-'V(l-|-f)==a mount due in 6 months, [and 

3. S+S(l+-r) + -S'(l+¥)= am oiiri t due in 9 months, 

4. 5+5(l+-J)+5 , (l+J r )+5(l+| r )= amount due 
in 1 year. [( l+-y> )=.$2556.25. 

5. .-. ^=$625-f$(;25(l+-o T 6) +$62 5(l+- 1 ¥ 2 )+$625- 

6. 100%=initial value. 

7. 6%=annuity. 

8. $2556.25=annuity. 

9. .-. 6%=$2556.25. 

10. 1%=4 of $2556.25=$426.0416|, and [value. 

111. 100%=100 times $426.0416f=$42604.16f=initial 



100%=initial value. 
l-^%=°i uar terly annuity. 
$625==quarterly annuity. 
... 1£%=$625. 

1%=JT of $625=$416.6666|, and 



G. 100%=100 times $416.6666-|=$41666.66J. 



100 



FINKEL'S SOLUTION BOOK. 



III. 



Initial value of A=$41979.16f, 
Initial value of B=$42604.16f , and 
Initial value of C=$41666.66|. 



(R. H. A., p. 310, prod. 5.) 
CASE II. 



Given 



Annuity, 

Interval, 

Rate, and 

Time the perpetuity is deferred, 

Let .S=the annuity, r=the rate, and R- 



to find the present 
value of a deferr- 
ed perpetuity. 

\-\-r. Then b}' Case 



L, the initial value of S is S-r-r. To find the present value of 

the initial value, we use formula III., compound interest. .'. P 

S S 

which t is the time the perpetuity 



_ r(l+r)« 
is deferred. 



R\R—\) 



I. Find the present value of a perpetuity of $250 a year, de- 
ferred 8 years, allowing 6% interest. 
By formula, 

P= „.,l .. = .. . ..J 250 .. - ^ J 250 .. =$2614.22. 



II.S 



III. 



R\R—\) (l+.06) 8 (l+.06— 1) .OGCl.OG) 1 
By 100% method. 
(1.) 100%=initial value. 
(2.) 6%=annuity. 
(3.) $250=annuity. 
(4.) .-. 6%=$250. 
(5.) 1%=J of$250=$41f,and 
(6.) 100%=100 times $41f=$4166.66f=initial value. 

'1. 100%=present value of $4166 66§ due in 8 years 
at 6%. 

2. 159.38481 %=( 1.06) 8 Xl00%=compound amount 
of the present value for 8 yr. at 6%. 

3. .-. 159.38481 %=$4166.66$, 
4 - 1 %=i59 3 1 848i of $4166.66f=$26.1422, and 
5. 100%=l'00 times $26.1422 = $2614.22 = present 

value. 

.-. The present value of a perpetuity of $250 a year de 
ferred 8 years at 6% interest=$2614.22. 

Find the present value of an estate which, in 5 years, is 
to pay $325 a year forever ; interest 8%, payable semi- 
annually. 

By formula, 

5 $325 $325 



(7.) 



$2690.67. 



[(1.04) 2 — 1](1.08) 5 .0816(1.08) 5 



ANNUITIES. 101 

By 100% method. 
(1.) 100%=initial value. 
(2.) 4%=amount due in 6 months. 

(3.) 4%-f(1.04)X4%=8.16%=amount due in 1 year. 
(4.) $325=amount due in 1 year. 
(5.) .-. 8. 16% =$325, 
.<!(6.) l%=gig of $325=$39.828431, and [value. 

(7.) 100%=100 times $39.828431= $3982.8431 = initial 

1. 100%=present value of $3982.8431. 

2. 146.93281 %=( 1.08) 5 Xl00%=compound amount 
of 100% for 5 yr. at 8%. 

l(8.)<[3. .'. 146.93281 %=$3982.8431, 

4. l%=nnnAreTT of $3982.8431=$26.9067, and 

5. 100% =100 times $26.9067 = $2690.67 = present 
value. 

III. .-. $2690. 67=present value of the estate. 

(R. H. A., p. 311, prod. 4.) 
Explanation. — The initial value is a sum of money which 
placed on interest at 8% payable semi-annually will produce 
$325 per year. But 8% payable semi-annually is the same as 
8.16% payable annually. Hence 8.16% is the annual payment. 
But $325 is the annual payment. Hence 8.16%=$325, from 
which we find that $3982.8431 is the initial value, or the amount 
that will produce $325 per year. Then the present value of a 
sum of money that will pay $325 is $3982.8431 if the payments 
are to begin at once, but $3982.843 l-f-( 1.08) 5 if the payments 
are not to begin until the end of 5 years. 

CASE III. 

{Rate, ^1 

™. , ^' , >to find the present valve of an an- 

lime to run, and ( \ . 

T . , nuity certain. 

Interval, J J 

(a) Let P denote the present value. The amount of P for n 
years=PR a =A. 

Let .S=the payment, or amount due the first year. 
►S-|-5Y?=the amount due the second year. 
6 , -|-6Y?-|-67i >2 =the amount due the third year. 
s4-^7?+S7? 2 +67? 3 = the amount due the fourth 
year. [due the ;zth year. 

tf-f SR + SR 2 +SR* + + 67^-! = amount 

.-. A=S+SX+SJt 2 +SJ?* + -\-SP»-i 

...(1) 

AR=SR+SR 2 +SR*+SR*+ SR n 

(2), by multiplying (1) by R. 
AR—A=SR a —S. . . (3), by subtracting (1) from (2). 

•'■ A== S ^Sl^ ( 4 ) But PRn = A - 



102 FINKEL'S SOLUTION BOOK. 

••' ^^= ^-1) "" (*>' Whence 
5(^-1) __S_ R^-j 

_/?»(A 5 — 1) ~i?— 1 X R» ^ 0)t 

(b.) When the annuity is to begin at a certain time, and then 
to continue a certain time. 

Let ^=the number of years the annuity is deferred, and q=^ 
the number of years the annuity continues. Then 

S R P+Q 1 

JP=— — - X — n~zn =the present value of an annuity S, for 

R — 1 JriP^Q. 

the time {p-\-q) years, and 

<? Dp 1 

P"=— — -x — ^— =the present value of an annuity S, for/ 

years. 

„ w n „ S Rp+Q—1 S 7?p— 1 S 

• p — p p — v V => 

" R— 1 A Rp+1 R—l^ JR* R—l 

V Rp+v R* } R^l\_ /?*+« K Rp'J R— lli?" 

)~ R— 1 A Rp+z K > 



Rp+q 



I. Find the present value of an annuity of $250, payable an- 
nually for 30 years at 5%. 
Given S, n, and r. 
By formula, 

P- S v*"- 1 J25Q (l.Q5)3Q-l _ 
^-^=T X ^^-^05" X (1.05)" -* 384 * 11 * 5 - 
By 100% method. 
(1.) 100 %=initial value. 
(2.) 5%=annuity. 
(3.) $250=annuity. 
(4.) .-.5% =$250, 
(5.) l%=\ of $250=$50, and 

(6.) 100%=100 times $50%=$5000=initial value of 
an immediate perpetuity of $250 per year. 

1. 100%=present value of an annuity deferred 30 
years. [ent value for 30 years. 

2. 432.19424%=(1.05) 30 Xl00%=amount of pres- 
(7 )J3. .-. 432.19424 %=$5000, 

4. i% = __J ¥¥ _ ¥ of $5000— $11.568865, and 

5. 100%=100 times $11.568865=$1156.8865=pres- 
ent value of annuity of $250 deferred 30 years. 

(8-) * .'. $5000— $1156.8865=$3843.1135=present value 
of an annuity continuing 30 years. 
III. V. $3843.1135=present value of an annuity of $250, payable 
annually for 30 years, 



II. 



ANNUITIES. 



103 



Remark. — Since $5000 is the initial value which, in this case, 
is also the present value of an immediate perpetual annuity, or 
perpetuity of $250, and $1156.8805 the present value of an an- 
nuity of $250 deferred 30 years, $5000— $1156.8865=$3843.1135= 
the present value of an annuity of $250 continuing for 30 years 
at 5%. 



I. 



Find the present value of an annuity of $826.50, to com- 
mence in 3 years and run 13 years, 9 months, interest 
6%, payable semi-annually. 

Given £=$826.50, r=.06, p=B years, and ?=13f years. 

When interest is payable semi-annually, i?=(l-|-|) 2 . 

By formula (7), 



II. 



P-- 



S 



7? 1 R ip+u) 



R*—l $826.50 (1.0609)^—1 



.0609 



X 



(1.0609) 



=$6324.69. 



By 100% method. 



(1.) 

(2.) 

(3.) 
(4.) 
(5.) 
(6.) 



(8.) 



(9.) 



(10.) 



100%=initial value. 
3%=amount due in 6 months. 
3%+3% (1.03)=6.09%=amount due in 1 year. 
$82650=amount due in 1 year. 
.-. 6.09%=$826.50, 

l%=T-ir of $826.50=$135.712643, and 
100% = 100 times $135.712643 = $13571.2643= 
initial value 

1. 100%=present value of a perpetuity of $826.50 

deferred 3 years. 

2. 119.40523%=(1.0609) 2 times 100%=amount of 

present value for 3 years. 

3. .-. 119 40523%=$13571.264: , >, 

4. l%= lliM i 0528 of $13571.2643=$113.6586, 

5. 100%=100 times $113.6586=$11365.86=present 

value of such a perpetuity deferred 3 years 

1. 100%=present value of such a perpetuity deferr- 

ed 16j vears. 

2. 269.212027%=(1.0609) lr,3 -i times 100% =amount 

of present value for ~[(V\ vears 

3. .-. 269.212027 %=$13571.2643, 

4. l% = & 6a . a ia o27 of $13571.2643=$50.4117, 

5. 100%=100 times $50.4117 = $5041.17 = present 

value of such a perpetuity deferred 16| years. 

.-. $11365.86— $5041. 17=$6324.69=present value 
of an annuity of $826.50 deferred 3 years and 
continuing - 13f years. 



III. /. $6324.69=present value of $826.50, etc. 



104 FINKEL'S SOLUTION BOOK. 

If the annuity is to begin in f years and continue forever, the 

formula, 

S Rv— 1 S 

P=l— — -X r, , becomes Jr. 



II. 



R— 1 ~ .#*+« ifr> ( i?— 1 ) ' 

For, since ^ = ^[ (l-^-j _ [l_i-] J if ^ =GO , the 
quantity 

1 — r^t =z1 — rp+^ ==1 ~^ ===1 ~°' a PP roaches 1 as its limit > 
and we have ^ = ^[(l_ ) -(l-^j ] =( -^-^. 

I. Find the present value of a perpetual annuity of $1000 to 
begin in 3 years, at 4% interest. 

By formula, [value of the annuity. 

By 100% method, 
f . (1.) 100%=initial value. 
(2.) 4%=annuity. 
(3.) .$1000=annuity. 
(4.) .-. 4%=$1000, 

(5.) l%=i of $1000=$250, and [$1000. 

(6.) 100%=100 times $250=$25000=initial value of 

1. 100%=present value. 

2. 112.4864% = (1.04) 3 times 100% = amount of 
present value for 3 years at 4%. 

L (7.)^ 3. .'. 112.4864 %=$25000, 

4. l%= TT 2i864 of $25000=$222.2492, and 

5. 100%=100 times $222.2492=$22224.92=present 
value. 

III. .'. $22224.92=present value of an annuity of $1000 to be- 
gin in 3 years at 4%. 

CASE IV. 

C Annuity, "^ 

I Rate ( 

Given ■< T . ' -, , >to find the final or forborne value, 
j Interval, and ( 

l^Time to run, J 

Let $=amount due first year. 

#-|-»S r i?=amount due second year. 

S-\-SR+SR 2 =amount due third year. 

S-\-SR+SR 2 -\-SR*=iimo\mt due the fourth year. 

S+SR+SR 2 +SRz + +3R*- 1 = amount due 

the nth year. 

Let .^4:=amount due the nth year. 

.-. A=S+SR-{-SR 2 +SRz+ +SR*- 1 ... (1). 



ANNUITIES. 



105 



AR=SR-}-SR 2 +SRZ-\-SR±+ +67? n 

. . (2), by multiplying (1) by R. [from (2). 

.-. AR-A=SR«— S (3), by subtracting (1) 

- ^W 1 w 

A pays $25 a year for tobacco ; how much better off would 
he have been in 40 years if he had invested it at 10% 
per annum? 

By formula, 

^=^ II X(^ n —l)=^X[(1.10) 40 —l] = $11064.8139. 

By 100% method. 

1. 100%=initial value. 

2. 10%=annuity. 

3. $25=annuitv. 

4. .-. 10%=$25, 
II.<! 5. 1%= T V of $25=$2.50, and 

6. 100%=100 times $2.50=$250=initial value. 

7. $44.2592556=[(1.10) 40 — l]X$l=compound interest of 
$1 for 40 yr. at 10%. [$250 for 40 yr. at 10%. 

8. .'. $11064.8139=44.2592556 X$250=compound int. of 
III. .-. He would be $11064.8139 better oft'. 

Remark. — $250 placed on interest at 10% will produce $25 
per year, [f this interest be put on interest at 10%, instead of 
spending it for tobacco, it will amount to $11064.8139 in 40 
years. This would be a very sensible and profitable investment 
for every young man to make, who is a slave to the pernicious 
habit. 



I. 



II. 



An annuity, at simple interest 6%, in 14 years, amounted 
to $116.76 ; what w T ould have been the difference, had it 
been at compound interest 6% ? 
(1.) 100%=initial value, or the principal that would 

produce the annuity. 
6%=annuity for 1 year. 
84%=14x6%=annuity for 14 years. 
'1. 100%=6%, 

2- 1%-rio of 6%=^%. and . [1 year. 

3. 6%=6 times -^ r %=2 9 ^%=interest on annuity for 

4. 32.76 %=91 times ^%=interest on annuitv for 
(1+2+3+ +14), or 91 years. 

84%+32.76%=116.76%=whole amount of the 

annuity. 
$116.76=whole amount of the annuity. 
/. 116.76%=$116.76, 
1 %=TTiT« of $116,76=$1, and 
100%=100 times $l==$100=initial value. 
6%=6 times $l=$6=annuity. 



(2.) 
(3.) 



(4.) 



(5.) 

(6.) 

(7.) 

(8.) 

(9.) 

(10.) 



106 



FINKEL'S SOLUTION BOOK. 



(11.) $1.260904=[(1.06) 14 — 1]X$1= compound inter- 
est on $1 for 14 yrs. at 6%. 
(12.) $126.0904=1.260904 X $100= compound interest 
on $100 for 14 yrs. at 6%. 
1(13.) ••• $126.0904— $116.76=$9.3304=difference. 
III. .-. The difference=:$9.3304. 



Given 



Solving P 



CASE V. 
Final Value or Present Value 
Rate, and 
Time to run, 

6 1 R»— 1 



R- 



X 



P(R— l)R l 



R»—l 

forborne value, 
S 



-1 
rPX 



R a 



> to find the annuity, 
with respect to S and we have 



R a — 1 
formula in 



. . . (1). If ^4=the final or 
the last case, we have A= 



R—l 



by the 

X^? n — I- Solving this with respect to S, we have. 

( R-l)A _ rA 

^--R^^-W^A {Z) - 

How much a year should I pay, to secure $15000 at the 

end of 17 >ears, interest 7% ? 
By formula (2), 

rA .07 X $15000 



:$486.38. 



IT. 



and 

=1428f 



dnitial value. 



5 R-—\ (1.07) 17 - 
By 100% method. 
(1.) 100%=annuity. 

(2.) 7%=annuity. 
(3.) .-. 7%=100%, 
(4.) l%=4of 100%=14f%, 
(5.) 100%=100 times 14f %■- 

"1. 100%=present value of 1428^-% due in 17 years. 
2. 315.8815%=amount of present value for 17 years. 
/A v 1 3. .'• 315.8815%=1428*%, 

^4. I^-htW^ 1428f-%=4.522591%, and 

5. 100%=100 times 4.522591 %=452.2591%=pres- 
ent value. 
(7.) s .*. 1428f-% —452.2591% = 976.3223% = present 

value of an annuity running 17 years. 
(8.) 3.1588152%=(1.07) 17 times l%=amount of 1% 

for 17 years. 
(9.) 3084.0217%=(1.07) 17 times 976.3223%=amount 
of 976.3223% for 17 years at 7%. 
(10.) $15000=amount, or final value. 
(11.) .". 3084.0217 %=$15000. 
(12.) 1%=30¥tV2TT of $15000=$4.8638, and 
( 13. ) 100 %=100 X $4.8638=$486.38=annuity . 



ANNUITIES. 107 

III. .-. I must pay $486.38. 

CASE VI. 

( Annuity, ) , c , ,. 

) t^ ; 17 i r 4-l a •*. j f to find time it 

Given < Present Value of the Annuity, and > 

( Kate, ) 

In formula (6), Case III., we have P=— — - X ' n — , whence 

R n ^ _P(R—1 1 __Pr_ 1 PrS—Pr 

~~R^ S ' or it^ -- ^' R~ S ~~ S~' 

s 

.•.R n =— — — - (1). Applying logarithms, 

n log. /?=log. [ s _ jPr ] • 

i r s 1 , r, loff- S— log. (S—Pr) , x 

I. In how many years can a debt of $1,000,000, drawing 
interest at 6%, be discharged by a sinking fund of 
$80,000 per year? 

By formula (2), 
_i og.S— lo g. ( ^— 7 Y) = 1og. 80000— log (80000— 1 000 00 X. 6) 

log. R log. 1.06 

log. 80000— log. 20000 _ 4.903000— 4.301030 __.002060_ 
log. 1.06 .025306 .025306 

years. 

By another method. 
Assume $1,000,000 to be the present value of an annuity of 
$80000 a year. Then $12.50 may be considered as the present 
value of $1 for the same time and rate. By reference to a table 
of present worths $12.50, which is lOOOOOO-y-SOOOO, will be 
found to be between 23 and 24 years. 

Note. — A table of present worths may be computed by form- 
ula (6.), Case III., in which put £=$1. 

I. In what time will a debt of $10000, drawing interest at 
6%, be paid by installments of $1000 a year. 
By formula, 
_ j Q g. S— log. ( S—/ y)_ log. 1000— log.( 1000— lOOOQx.06) 
log. R log. 1.06 

3—2.602060 1r _ _ _ Q1 , 

= — ~._ t , /A/ , =1d./2;j years=lo yr. 8 mo. 21 da. 
•02o30b 

By another method. 

Assume $10000 to be the present value of an annuity of $1000 

a year. Then $10000-r-1000=$10=the present value"' of $1 for 

the same time and rate. By referring to a table of present worth 

we find this amount between 15 and 16 years. ,\ The time is 15 

years -\- 



108 FINKEL'S SOLUTION BOOK. 

The compound amount of $10000 for 15 yr. at 6%= $23965.58 
The final value of $1000 for 15 years at 6%= $23275.97 

Balances $ 689.61 
This balance, $689.61, will require a fraction of a year to dis- 
charge it. The part of a year required, will be such a fraction of 
a year as the amount of $689.61 for the fraction of a year is of 
$1000. 

6% of $689.61 for the fraction of a year=$41.3766x fraction 
of a year. 

.-. $689. 61-f-$41. 3766 X fraction of a year=the amount of 
$689.61 for the fraction of a year. This amount divided by 
$1000, a yearly payment, will give the fraction. 
.61+$41.3766X/W/*W . . 
$1000 =fraction, whence 

$Q89M+U1.37 66Xfraction=$lOOOXfracti'on 

.'. $1000 Xfraction— $41. 3766 Xfraction=$Q89M, or 

.-. $958,628 X/™^'^==$689.61. 

- ,. 689.61 Q ., 1Q , 

.•.Jractwn=-———~=8 months, 19 days. 

95o.ozo 

.-. The whole time=15 yr. 8 mon. 19 da. 

CASE VII. 

C Annuity, ) 

Given 1 Time to Run, and > to find the rate of in- 

( Present Value of an Annuity, ) terest. 

From the formula (6), Case III, P=~b — :X — ^~ » we obtain 

R* 1 p 

— -—-=— .... (1). This is the simplest expression we can ob- 
tain for the rate as the equation is of the nth degree and can not 
be solved in a general manner. 

I. If an immediate annuity of $80, running 14 yr., sells for 
$650, what is the rate? 

By formula, 

i? n — 1 P $650 Q10K 

-r^=s=m= s - 12 ^ or 

-7ZT-, — rrr=8.125. Solving this equation by the method of 

r (1+r) 14 6 • . .m J 

Double Position, we find r=8%-|-. 
By another method. 
$650-^$80=8.125. By referring to a table of present worths 
of $1, corresponding to 14 years, we find it to be between 8 and 
9%. 



PROBLEMS. ]09 



PROBLEMS. 

1. What is the amount of an annuity of $1000, forborne 
15 years, at 3-J-% compound interest? Ans. $19295.125 

2. What will an annuity of $30 payable semi-annually, 
amount to, in arrears 3 years at 7% compound interest? 

Ans. 



3. What is the present worth of an annuity of $500 to con- 
tinue 40 years at 7% ? Ans. 



4. What is the present worth of an annuity of $200, for 7 
years, at 5% ? Ans. $1152.27. 

5. A father presents to his daughter, for 8 years, a rental of 
$70 per annum, payable yearly, and the reversion for 12 years 
succeeding to his son. What is the present value of the gift to 
his son, allowing 4% compound interest? Ans. 

6. A yearly pension which has been forborne for years, at 
6%, amounts to $279 ; what was the pension? Ans. $480.03. 

7. A perpetual annuity of $100 a year is sold for $2000 ; at 
what rate is the interest reckoned? . Ins. 

8. A perpetual annuity of $1000 beginning at the end of 10 
years, is to be purchased. If interest is reckoned at 3^%, what 
should be paid for it? Ans. 

9. If a clergyman's salary of $700 per annum is 6 years in ar- 
rears, how much is due, allowing compound interest at (\ c / ? 

Ans. $4882.72. 

10. A soldier's pension of $350 per annum is 5 years in ar- 
rears; allowing 5% compound interest, what is due him? 

Ans. $1933.97. 

11. What annual payment will meet principal and interest of 
a debt of $2000 due in 4 year a 8% compound interest? Ans. — 

12. What is the present worth of a perpetual annuity of$(J00 
at 6% per annum? Ans. $10000. 

13. What is the present value of an annuity of $1000, to com- 
mence at the end of 15 years, and continue forever, at (\ ( / per 
annum? Ans. $6954.40. 

14. To what sum will an annuity of $120 for 20 years amount 
at 6% per annum? Ans. $4414.27. 

15. A debt of $8000 at 6% compound interest, is discharged 
by eight equal annual installments; what was the annual install- 
ment'? Ans. $1288.286. 



110 



FINKEL'S SOLUTION BOOK. 



CHAPTER XV. 

MISCELLANEOUS PROBLEMS, 

INVOLVING THE VARIOUS APPLICATIONS OF PERCENTAGE. 

I. Sold a cow for $25, losing 16f% ; b >ught another and sold 
it at a gain of 16%; I neither gained nor lost on the two ; what 
was the cost of each? 



II. 



r A. 



100%=cost of the first cow. 

16f%=loss. 

100%— 16|%=83i%=selling price. 

$25=selling price. 



83^%=$25 : 
1 



and 



B. 



1%=~ of$25=$.30, 

100%=100 times $.30=$30=cost of first cow. 

$30 — $25=$5, loss on the first cow, and gain on 

second* cow. 
100%=cost of second cow. 
16%=gain. 

$5=gain. 
.-. 16%=$5. 

5. 1%= T V of $5=$.3125, and [cow. 

6. 100%=100 times $.3125=$31.25=cost of second 
$30=cost of first cow, and 
$31.25=cost of second cow. 

Remark. — Since I lost $5 on the first cow, and neither gained 
nor lost on the two, I must have gained $5 on the second cow. 
.-. 16%=$5. 

I. There have been two equal annual payments on a 6% note 
of $175, given 2 years ago this day. The balance is 
$154.40; what was each payment? 



III. 



II. 



(!■) 

(2.) 
(3.) 
(4.) 
(5.) 

(6.) 



(7.) 



100 %=a payment. 
100% =$175, 

1%=^ of $175=$1.75, and 
6%=6 times $1.75=$10.50=interest for 1 year. 
$175-j-$10. 50= $185.50 = amount before paying 
the payment. [payment. 

$185.50— 100% =amount left after paying the 
100%=$185.50— 100% , 

1 %=rhi of ($185.50— 100% )=$1.855—1%, and 
6%=6 times ($1.855— 6%)=$11.13— 6%=inter- 

est for second year. 
$185.50— 100%+$11.13— 6%=$196.63— 106% = 

amount before paying the last payment. 
$196.63 — 106% — 100% = $196.63 — 206% = 

amount left after paying the last payment. 



MISCELLANEOUS PROBLEMS. 



Ill 



(8.) $154. 40=amount after paying the last payment 
(9.) .-• $154.40=$196.63— 206%. 
(10.) 206%=$196.63— $154.40=$42.23, 
(11.) l%=jh of $42.23=$.205, and 
1^(12.) 100%=100 times $.205=$20.50=the payment. 

III. .'. $20.50=lhe payment. 

Re?nark. — In this solution we are obliged (o use the minus 
sign, — , which is no obstacle to the student of algebra, but to the 
student of arithmetic it may seem insurmountable. To avoid 
this sign, we give another solution. 



II. 



(1) 

(2.) 

(3.) 

(4-) 
(5.) 

(6.) 

(V.) 
(8.) 



(9.) 



(10.) 



(11.) 
(12.) 
(13.) 

(14.) 
(15.) 



100%=thc payment. Then 
$154.40+100%=amount of the debt at the end of 

of the second year. 
•100%=principal that produced this amount. 
6%=interest. 
106%=amount. 

... 106%=$154.40+100%, [and 

l%==Tk of ($154.40+100% )=$1.4666,VH**» 
100%=100 times ( $1.4566 & \\ %% % ) = $145.66^ 

-j-94-i--|%=a mount at end of the first year after 

paying off the payment. 
$145.66A+94^%+100%= $1 45.66^ + 194-1 { % 

=a mount before paying off the payment = 

amount at end of first year. 
100%=the principal that produced it. 
()%=interest. 
106%=amount. 
.-. L06%=$145.66A+194if%, 

l%=r^of($145.66 T \+194if%) = $1.37^HJ + 
1.83**%, and 

100% = 100 times (*1.37«H + l- s; >-. : 'sVV%) = 

$137ttfJ-+ 1 83^V%= the amount at first. 
$175=the amount at first. 
/. $137HU+183* 5 T ft r %=$175. 
183A%%=$37M» 



1%=$37«^-^183^\=$.205 J and 

100% =100 times $.205=$20.50=the payment. 
III." .-. $20.50=the payment. (7?. H. A., p. 26^prob. 5.) 

Explanation. — $154.40=the amount after paying off the last 
payment. .-. $154.40-f-100%=amount before paying of the last 
payment, or it equals the debt at the end of the first year plus 
the interest on this debt for the second year. .-. We let 100 r / = 
the debt at the end of the first year, 106%=amount of 100$ for 
1 year. .'. 106% =$154.40 + 100%. Then proceed as in the 
solution. 

I. If a merchant sells f of an article for what £ of it cost, 
what is his gain % ? 



112 



FINKEL'S SOLUTION BOOK. 



1. 100%=cost of whole article. 

2. 87i%=J of 100%=cost of J cf the article. 

3. 87-^%= se ll m g price of f of the article. 
II.<|4. 29^%=} of 87|%?=selling price of | of the article. 

5. 116f%=4 times 29^%= selling price- of the whole 

ja T*1~1 (~* 1 f* 

6. .-. 116f %— 100%=16f%=gain. 

III. .-. 16f%=his gain. (Milne's Prac, p. 360, prod. 51.) 

I. A merchant sold goods to a certain amount, on a commis- 
sion of 4%, and having remitted the net proceeds to the 
owner, received ^% for prompt payment, which 
amounted to $15.60. What was his commission? 

(1.) 100%=cost of goods. 

(2.) 4%=commission. 

(3.) 100%— 4%=96%=net proceeds. 

rl. ^%=amount received for prompt payment. 

| 2. $15.60=amount received for prompt pavment. 

(4.W3. .-. ±%=$15.60. 

II.<! 4. 1%=4 times $15.60=$62.40. 

[5. 100%=100 times $62.40=$6240=net proceeds. 

(5.) .-. 96%=$6240. 

(6.) l%=9 1 e of $6240=$65, and 

(7.) 100% = 100 times $65=$6500=cost of goods. 

rl. 100% =$6500. 

(8.)<|2. l%= T |o of $6500=$65, and 

13. 4% =4 times $65=$260=his commission. 
III. .-. His commission=$260. 

( Greenleaf's JV. A., p. ^l.prob. 11.) 



I. If I sell 30 yards of cloth for $132, and gain 10%, how 
ought I to sell it a yard to lose 25% ? 

$132=selling price of 30 yards. 
$4.40=$132-^30=selling price of one yard. 
100%=cost of one yard. 
10%=gain. 

100%+10%=110%=selling price per yard. 
$4.40=selling price per yard. 
.-. 110%=$4.40. 
1%=tto of $4.40=$.04, 
100% =100 times $.04=$4=cost per yard. 
100% =$4. 

1%=t*tt of$4=$.04, 
25% =25 times $.04=$l=loss. 
.-. $4 — $l=$3=selling price per yard to lose 25%. 
I must sell it at $3 per yard to lose 25%. 

{Stoddard's Complete, p. 206,prob. 9.) 




III. 



MISCELLANEOUS PROBLEMS. 113 

A merchant receives on commission three kinds of flour ; 
from A he receives 20 barrels, from B 25 barrels, and 
from C 40 barrels. He finds that A's flour is 10% better 
than B's, and that B's is 20% better than C's. He sells 
the whole at $6 per barrel. What in justice should 
each man receive? 

(1.) $6=selliii£ price of 1 barrel. 

(2.) $5-10=sel1ing price of (20+25+40), or 85 barrels. 

(3.) 100%=value of C's flour per barrel. 

(4.) 120%=value of B's flour per barrel. 

rl. 100%=120%. 
(5.) 2. l%=jfa otl20%=li%, 

13. 10%=10 times l-l%=12%. 
(6.) 120%+12%=132%=value of A's flour per barrel. 
(7.) 4000% =40 times 100%=wbat C received. 
IV (8.) 3000%=25 times 120%=what B received. 
f9.) 2640%=20 times 132%=what A received. 
(10.) 9640%=4000%+3000%+2640%=what all ree'd. 
(11.) $5 10= what all received. 
(12.) .-. 9640%=$510. 

(13.) 1%=W^ of $510=$.52fJ|, and [received. 

(14.) 4000% =4000 times $.52-^ = $211^=what C 

(15.) 3000% =3000 times $.52f^ = $158£|-J=what B 

received. [received. 

1(16.) 2640%=2640 times $.52fi|=$139^=what A 

( $139if}=A's share, 
III. .-. < $15SiJf-=B\s share, and 
( $21H+'=C's share. 

( Greenlcafs National Arith. p. ^2.) 

I. |- of B's money equals A's money. What % is A's money 
less than B's, and what % is B's money more than A's? 

1. 1()0%=B\s money. 

2. 75% =} of 100%=A's money. 

-r-r J 3. 100 '/( — 75$ =25$ =excess of B's money over A's. 
U -U. 75%=100% of itself, 

1 %>=J, of 10()%=H% , and [than A's. 

25%=25 times l-J-%=33^%=the % B's money is more 

A's money is 25% less than B's, and 

B's money is 33^-% more than A's money. 

(Stod. Comp.,p.203, prob. 19.) 



6. 



Ill 



T. At what price must an article which cost 30 cents be 
marked, to allow a discount of 12^% and yield a net 
profit of 161 % ? 



114 



FINKEL'S SOLUTION BOOK. 



II. 



III. 



(1.) 100%=30/, 
(2.) l%= T io of 30/= T V, and 
(3.) 16f %=16f times ^ r /=5/=profit. 
(4.) 80/-|-5/=35/=the price at which it must sell to 
gain 16f%. 
'1. 100%=marked price. 

2. 12^%=discount from marked price. 

3. 100%— 12|-%=87i%=selling price. 

4. 35/— selling price. 

5. .-. 87i%=35/. 



(5.) 



6. 



^=87i ° f35 - 



.40/, and 
7. 100%=100 times .40/=40/=marked price. 
.*. 40^=marked price. 



( Seymour 's Prac, p. 203, prob. If. 



II. 



111. 



Had an article cost 10% less, the number of % gain would 
have been 15% more ; what was the gain? 

1. i(%)%=selling price. 

2. 100%=actual cost price. 

3. 100%— 100%=gain. 

4. 100% — 10%=90%=supposed cost. 

5. 100% — 90%=conditional gain. 

6. 90%=100% of itself. 

7. l%=9V°f 100%=H%. 

8. 100%— 90%={100— 90) times 11% 

=conditional gain %: 

9. /. \° X100 %— 100%— (100% — 100%) = ±X100% = 

10. 15%=difference. 

11. .'.%XlOO%=lb%. [the actual cost. 

12. 100%=§ times 15%=135%=selling price in terms of 

13. .'. 135%— 100%=35%=gain. 

(R. H. A., p. lf.0G.prob. 87.) 



\fxioo%— 100% 

[difference. 



, 35%=gain. 
A literal solution. 
Let 6 , =selling price and C=the cost. Then S — C=gain ;md 

g Q $ $L (J 

— — ^ - =rate of gain. S — T 9 ^C=conditional gain and 



1_0 ^ c 

-=conditional ra 



- 9 -C 

i <~ 



C 



te of 



gain. 



$S=±fo C, whence S==f£C=1.35 C. . 
.-. Rate of gain=.35C-r-C=.35=35%. 



\°S—C S—C 
C C~~ 

1.35 C— C=.35C=jrain 



A' or 



I. In the erection of my house I paid three times as much for 
material as for labor. Had I paid 6% more for labor, 
and 10% more for material, my house would have cost 
$3052. What did it cost me? 



MISCELLANEOUS PROBLEMS. 



115 



II. 



III. 



(2.) 

(3.)- 



(4.)f, 
14. 
(5.) 
(6.) 

(7.) 

(8.) 

(9.) 
(10.) 
(11.) 

.-. $2800^ 



100%=cost of labor. 

300% =3 times 100%=cost of material. 

100%=100%, 

1%=1%, and 

6%=6%. 

100%+6%=106%=supposed cost of labor. 

100%— 300%, 

l%=To-ir of300%=3%, and 

10%=10 times 3%=30%. 

300%+30%=330%=supposed cost of material. 

330%-f-106%— 436%==supposed cost of house. 

$3052=sup posed cost of house. 

.-.436% =$3052, 

1%= T ^ of $3052=$7, and 

100%=100 times $7=$700=cost of labor. 

300%=300 times $7=$2100=cost of material. 

$2100+$700=$2800=cost of house. 

=cost of the house. 



I. 



II, 



I invest § as much in 8% canal stock at 104%., as in (>% 
gas stock at 117% ; if my income from both is $1200, 
how much did I pay for each, and what was the income 
from each ? 

(1.) 100%=investment in gas stock. Then 

(2.) ()()|%=investment in canal stock. 

1. 100%;=par value of the gas stock. 

2. 117%=market value of the gas stock. 

3. .-. 117 9&=100%, from (1), 

4. 1%= T 1 T of 100%Hr£# , and 

5. 100%=100 times }0«%=85|f%==par value in 
terms of the investment. 

100%=85tf%, 

1%=-Hf%,and 

6%=6 times ^T%=5-^%=income of gas stock. 

l()()%=par value of canal stock. 

1(U%— market value. 

.-. 104%=66f%, 

l%=ri* of 6()|% = |;}%, and 

100%=100 times ff%=B4^%. 

10()%-=U4,\%, 

1 %=to l o of 64A%Hri% , ^d 

8%=8 times f, ;',%=") .^%=income of canal stock. 

5 ¥ \%+5^V=10^%=income from both. 

$1200=income from both. 

... 10^%==$1200, 

(10.) 1 '/<■=- — of $1200=$117, and 



(3.) 



r 1 - 

(5.) 1 2. 
1. 



(60 



(7- 
8. 
9. 



•is 



116 FINKEL'S SOLUTION BOOK. 

(11.) 100%=100 times $117=$1170G=investment in 

gas stock. [canal stock. 

(12.) 66f%=66f times $117=$7800 = investment in 

1(13.) 5^%=5A times $H7=$600=income from each. 

( $600=income from each. 
III. .'. < $11700=in vestment in gas stock, and 
( $7800=investment in canal stock. 

I. A man bought two horses for $300; he sold them for $250 
apiece. The gain on one was 5% more than on the 
other; what was the gain on each? 

1. $300=cost of both. 

2. $500=$250+$250=selling price of both. 

3. $500— $300=$200=gain on both. 

4. 100%=gain on first horse. Then 

5. 105%=gain on second horse. 

6. 205%=100%+105%=gain on both. 

7. $200=gain on both. 

8. .-. 205%=$200. 

9. 1%=Ht of $200=$f£, and 

10. 100%=100 times $f «=$97.56 T 4 T =gain on the first. 

11. 105%=105 times $fj=$102.43fj^=gain on the second. 

TTT $ $97-56 ¥ 4 T =gain on the first, and 

"'■ ( $102.43£f=gain on the second. 

Note. — In this solution, it is assumed that the gain on one was 
5% of the gain on the other more than the other, and this is the 
usual assumption. But the problem really means that the per 
cent, of gain on one, computed on its cost, was 5% more than 
the per cent, of gain on the other, computed on its cost. By this 
assumption, the problem is algebraic. The following is the 
solution: Let *=the cost of the first horse", and $300 — x, the 
cost of the second. Then $250 — #=gain on nrst, and $250 — 
($300— x)=x— $50, the gain on the second. ($250— x)-r-x= 
rate of gain on the first, and (x — $50)-r-($300 — *), the rate of 
gain on the second. Then (250 — x)~-x — (x — 50)-S-(300 — x)= 
-^L. Whence, by clearing of fractions, transposing and, combin- 
ing, x 2 — 10300 *=— 1500000, *=5150J = 5(VIo5b9=$147.7755, 
the cost of the first horse. $300— *=$152.2245, the cost of the 
second horse. $250 — #=$102.2245, gain on the first horse, and 
x — $50=$97-7755, the gain on the second horse. 

I. An agent sells produce at 2% commission, invests the 
proceeds in flour at 3% commission; his whole commis- 
sion was $75. How many barrels of flour did he buy 
at $5 per barrel? 



MISCELLANEOUS PROBLEMS. 



11! 



(1) 
(2.) 
(3.) 



(4.) 



II. 



III. 
I. 



(5.) 

(6.) 

(8.) 

(9.) 

(10.) 

(11.) 
c (12.) 



100%=value of the produce. 

2%=the commission. [vested in the flour. 

100% — 2%=98%=net proceeds, or amount in- 

100%=cost of the flour. 

3 %=com mission on flour. 

100%+3%=103%=whole cost of the flour. 

.-. 103%=98%, 

1 %=tt>h of 98%= T 9 A% , and 

100%=100X T 9 A% = 95 T W%=cost of flour in 

terms of the value of the produce. 
98% — 95^%=2 T ^%=commission on flour. 

2%+2yV^%— 4t<T3"%— wno ^ e commission. 

$75=whole commission. 

•'• 4A 8 3%=$75, 

1%=$75-v-4 t W=$15.45, and [produce. 

100%=100 times $15.45=$1545 = value of the 

95 T W%==95 T VV time s $15.45= $1470 = value of 

the the flour. 
$5=cost of 1 barrel. 
$1470=cost of 1470-4-5, oi 294 barrels. 



The agent bought 294 barrels of flour. 

A distiller sold his whisky, losing 4% ; keeping $18 of 
the proceeds, he gave the remainder to an agent to buy 
rye at 8% commission; he lost in all $32 ; what was the 
whisky worth? 



(i-) 

(2.) 
(3.) 
(4.) 



(5.) 



100%=value of the whisky. 

4%=loss. 

100%— 4%=96%=amount he had left. 

96% — $18=amount he invested in rye. 

1. 100%=cost of the rye. 

2. 8 %=com mission on the rye. 

3. 100%+8%=108%=whole cost of rye. 



II ^ 



7. 



lt)8%=96%— $18, 

l%=rk of(96%-$18)= 

100%=100 times (f%- 

=cost of rye. 



=5-%— M6ti a nd 
-$.16f)=88f%— $16.66f 



(f%-$.16»)=7i% 



(6.) 

(V-) 
(8.) 
(9.) 

(10.) 

(11.) 



-$1.33J 



$1.33i=com- 
wholeloss. 



111. 



;oO: 



8%=8 times 

mission on rye. 
4%+(7i%-$1.33i)=lli% 

$32= whole loss. 

••• 1H%— $1.33t=$32 

lH%=$33.33i, 

1%=^ of $33.33-i=$3, and 

100%=100 times $3=$300=value of the whisky 
=value of the whisky. 

(R. H. A.,p.I,.06,prob9L] 



118 FINKEL'S SOLUTION BOOK. 

I. What will be the cost in New Orleans of a draft on New 
York, payable 60 days after sight, tor $5000, exchange 
being at \\ c /o premium? 

1. 100%=face of the draft. 

2. 1-^%— premium. 

3. 100%+l-J-%==10H%=rate of exchange. 

4. 5%=discount for one year. 
II.<! 5. |%=^ 6 g% of 5%=discount for 63 days. 

6. .-. 10H%— J%=100£%=co8t of the draft 

7. 100% =$5000. 

8. 1%= T ^ of $5000=$50, and 
19. lOOf %=100| times $50=$5031.25=cost of the draft. 

III. .'. $5031.25=cost of the draft. 

Explanation. — Observe that since the draft is not to be paid in 
New York for 63 days, the banker in New Orleans, who has the 
use of the money for that time allows the drawer discount on the 
face of the draft for that time, which goes (1) towards reducing 
the premium if there be any, and (2) towards reducing the face 
of the draft. 

Note. — The rate of exchange between two places or countries 
depends upon the course of trade. Suppose the trade between 
New York and New Orleans is such that New York owes New 
Orleans $10,250,000 and New Orleans owes New York $13,000,- 
000. There is a "balance of trade" of $2,750,000 against New 
Orleans and in favor of New York. Hence, the demand in New 
Orleans for drafts on New York is greater than the demand in 
New York for drafts on New Orleans and, therefore, the drafts 
are at a premium in New Orleans. But if New York owes New 
Orleans $13,000,000 and New Orleans owes New York $10,250,- 
000, the "balance of trade," $2,750,000, is against New York and 
in/awr of New Orleans. Hence, the demand in New Orleans 
for drafts on New York is less than the demand in New York 
for drafts on New Orleans and, therefore, the drafts are at a dis- 
count in New Orleans. 

If the trade between the two places is the same, the rate of ex- 
change is at par. 

The reason why the banks in New York should charge a pre- 
mium, when the balance of trade is against them, is that they 
must be at the expense of actually sending money to the New 
Orleans banks or be charged interest on their unpaid balance ; 
the reason why the New Orleans banks will sell at a discount is 
that they are willing to sell for less than the face of a draft in 
order to get the money owed them in New York immediately. 

Exchange is charged from ^ to -J%, and is designed to cover 
the cost of transporting the funds from one place to another. 



MISCELLANEOUS PROBLEMS. 



119 



II. 



III. 
I. 



What will a 30 days' draft on New Orleans for $7216.85 
cost, at -§% discount, interest 6% ? 

1. 100%=face of draft. 

2. f%=discount. 

3. 100%— | %=99f %=face less the discount. 

4. 6%=bank discount for 1 year. 

5. H%=imr of 6%=bank discount for 33 days. 

6. 99|%— H%=99&%= cost of the draft. 

7. 100%=$7216.85, 

8- 1%=t1tt> of $7216.85=$72.1685, and 

9. 99^^ =9 9^ t i mes $72.1685=$7150.094=cost of the 

draft. 
.-. $7150.094=cost of the draft. 
The aggregate face value of two notes is $761.70 and each 
has 1 year 3 months to run; one of the notes I had dis- 
counted at 10% true discount and the other at 10% 
bank discount, and realized . from both notes $671-50. 
Find the face value of both notes. 

100%=face of note discounted at bank discount. 
$761.70 — 100%=face of note discounted at true 

discount. 
10%=bank discount for 1 year. 
12^%=bank discount for 1 year 3 months. 

1. 100%=present worth of second note. 

2. 10%=interest on present worth for 1 year. 

3. 12^%=interest for 1 year 3 months. 

4. 100%+12£%=112|%=amount of present worth. 

5. $761.70— 100%=amount of the present worth. 

6. .-. 112i%=$761.70— 100%, 



II.S 



(1.) 

(2.) 

(3.) 

(4.) 



(5.) 



1%=: 



of ($761.70- 



8. 



(6.) 

(7.) 
(8.) 
(9.) 

(10.) 

(11.) 
(12.) 

(13.) 



-100%)=$6. 


7706£— f%, 


*06f- 


-t%) = 


= $677.06f— 


06S- 


-88$%) = 


= $84.63i — 



III. 



112£ 

100%=100 times ($6.77 

88f%=present worth. 
$761.70— 100%— ($677. 

ll^-%=true discount. [discount. 

$84.6*3i— ll£%+12£%==$84.63i+ 1 ,v% = whole 
$761.70— $671.50=$90.20=whole discount. 

$84.63i+l T V%=$90.20, 

1 %=-!- of $5.56f =$4,008, and 

Irs 
100%==100 times $4.008==$400.80=face of note 

discounted at bank discount. 
$761.70— 100%=$761.70— $400.80=$360.9O=face 
of note discounted at true discount. 
$400.80=face of note discounted at bank discount, and 
90==face of note discounted at true discount. 



hV/c 



i $400.* 
' ' I $360.? 



IU 



120 . FINKEL'S SOLUTION BOOK. 

1. A merchant sold part of his goods at a profit of 20%, and 
the remainder at a loss of 11%. His goods cost $1000 
and his gain was $100; how mnch was sold at a profit? 
(1.) 100%=cost of goods sold at a profit. Then 
(2.) $1000— 100%=cost of goods sold at a loss. 
(3.) 20%=profit on 100%, the part sold at a profit. 

rl. 100% =$1000— 100%. 
lL J 2. l%=r^of ($1000— 100% )=$10— 1%, 
W ]3. 11%=11 times ($10— 1%)=$110— ll%=loss on 

I the remainder. 

(5.) .'. 20%— ($110— 11%)=31%— $110=gain. 
(6.) $100=gain. 
(7.) -\ 31%— $110=$ 100. 
(8.) 31%=$210, 

(9. ) 1 %=i T of $210=$6ff , [profit. 

[(10.) 100%=100 times $6ff=677.41$f=part sold at a 
III. .-. $677.41-ff=value of the part sold at a profit. 

I. By discounting a note at 20% per annum, I get 22-£% 
per annum interest; how long does the note? 

1. 22|% of the proceeds=20% of the face of the note. 

2. 1% of the proceeds=— T of 20%=f% of the face of the 
note. ll % 

3. 100% of the proceeds=100 timss f%=88|- % of the face 
of the note. 

4. 100%=face of the note. 

5. 88f%=proceeds. 

6. 100%— 88f %=lli%=discount for a certain time. 

7. 20%=discount for 360 days. 

8. l%=discount for -^ of 360 days, or 18 days. 

9. ll^%=discount for 11^ times 18 days, or 200 days. 
III. .-. The note was discounted for 200 days. 

I. A man bought a farm for $5000, agreeing to pay princi- 
pal and interest in 5 equal annual installments. What 
will be the annual payment including interest at 6% ? 

1. 100%=one annual payment. 

2. .*. 100%=amount paid at end of the fifth year 
since the debt was then discharged. 

3. 100%=principal that drew interest the fifth year. 

4. 6%=interest on this principal. 

5. .-. 100%+6%=106%=amount of this principal. 

6. .*. 106%=100%=the annual payment. 
7- 1%= tU of 100%=|f% , and 

8. 100%=100 times ■§£ %=94!f %=principal at the 
beginning of the fifth year. 

9. 94£f%+100%=194|f%=amount at the end of 
the fourth year. 



II. 



(1.) 



MISCELLANEOUS PROBLEMS. 



121 



II. 



(2-) 



(3.) 



(4-) 



(5.) 



(6.) 

(<■) 
(8.) 
(9.) 



100%=principal at the beginning of the fourth 

year. 
6%=interest on this principal. 
100%+6%=106%=amount. 
.-. 106%=194frf%, 

1%=t*f ofl94U%=l-83A%%, ^d 
100%=100 times i:83JJftV%=183A%%=princi- 

pal at the beginning of the fourth year. 

183^9 %+100%=283A 5 iT9 %= amount at the end 
of the third year. 

100%=principal at the beginning of the third 

year. 
6%=interest. [third year. 

100%+6%=106% = amount at the end of the 
.-. 106%=283^fe%, 

l%=rkof283AW r %=2.67T¥iflAV%, and 
100%=100 times 2.67 t 4 AWt% = 267 t \WtV% = 

principal at the beginning of third year. 

267^frVr%+100% = 367AWrV% = amount at 

the end of second year. 
100%=principal at the beginning of second year. 
6%=interest. [year. 

100%+6%==106%=amount at the end of second 
.-. 106%=367 T VAVr%» 
5- l%=rk of 367iWWr%=3.46HH*W%, and 
6. 100%=:100 times 3.46^1|f|f %=346^§f Jif %= 
principal at the beginning of the second year. 

7- 346^!H%+100%=446nt!ttf%=amount at 
the end of first year. 

1. 100%=principal at the beginning of the first 

year, or the cost of farm. 

6%=interest. 

100%+6%=106%==amount at end of first year. 



ft 

2. 
3. 
4. 
5. 
6. 

7. 



106%=446ifmii% 



10/, 1 n f zLlfiiO 'J 8 5 7 4 0/ 191 98 8 5 244 7 0/ o nr j 

ITT yg-g- OI 44t) T8lU)48 1 7^ 4 ~ - 1 41 81 5T9 H ; A > ailCl 

100^/, 100 tinips4 91 98852447 oL A.') 1 9 8 8 5 2 4 47 

IUU70 — 1UU limes 4-^l*TTrTT4~T5/0 — 4Z1 4T8T9TT9¥ 

%=cost of the farm. 
$5000=cost of the farm. 
••• 421*ttHMfrJfc=W000, 
1 %=$5000-r-421^%^^=$lt8698+ J and 
100%=100 times $11.8698=$118G.98-f-=the an- 
nual payment. 



III. .-. $1180.1)8+3 



:the annual payment. 

(Milne's Prac. 



p. 861, f rob. 63.) 

-% 
I. A and B have $4700 ; fT"% of A ' s share equals '|^7% of 

"2 1° 
B's share; how much has each? 



122 FINKEL'S SOLUTION BOOK. 



!* ol„ =121 % ^-_3 _ of„ — 3 . 



IIJ 



T 70 — — 70 — TOU- 70 — ¥ o o U o" 
z - 2/0 — —70—20070 — 2-onnro 

%% _ 3 

3. .-.4% - 4¥ r° 70=H%. 



2 ^ 2 



4 -±L o1„ — —--—oL 2000/ 200 2 

q 0/ 70 — ^r^ A — ~ 3 ~ 70 — 300 — 3' 

d /° T0"0 

5- 60%= T %=|. 



2 



. b* 



s, 



III. 



6..,^%=| % = li% . 

7. .-. 1|% of A's==ll% of B ; s, 

8. 1% of A's=jy of H%=H % of B's, and 

9. 100% of A's=100 times |^%=74^% of B's. 

10. 100%=B's share. 

11. 74vV70=A's share. 

12. 106%+74^ r %==174 li 2 T %=sum of their shares. 

13. $4700=sum of their shares. 

14. .-. 174^%=$4700, 

15. 1%=^j of $470O=$27, and 

16. 100%=100 times $27=$2700=B's share. 

17. 74^ r %=74^ 7 - times $27=$2000=A's share. 
$2700=B's share and 
$2000= A 's share. 



CHAPTER XVI. 

RATIO AND PROPORTION. 

1, JEtcitio is the relative magnitude of one quantity as com- 
pared with another of the same kind; thus, the ratio of 12 apples 
to 4 apples is 3. 

The first quanity, 12 apples, is called the Antecedent , and the 
second quantity, 4 apples, the consequent. Taken together they 
are called Terms of the ratio, or a Couplet. 

2. TJie Sigtl of ratio is the colon, :, the common sign of 
division with the horizontal line omitted. 

Note. — Olney says, "There is a common notion among us, that the French 
express a ratio by divding the consequent by the antecedent, while the En- 
glish express it by dividing the antecedent by the consequent. Such is not 
the fact. French, German, and English writers agree in the above defini- 
tion. In fact, the Germans very generally use the sign : instead of -j-; and 



PROPORTION. 123 

by all, the two signs are used as exact equivalents." Some writers, 
however, divide the consequent by the antecedent, as a : b= — This is ac- 
cording to Webster's definition and illustration. To my mind, to divide 
the antecedent by the consequent is more simple and philosophical and 
should be universally adopted by all writers on mathematics. 

3. A Direct Ratio is the quotient of the antecedent di- 
vided by the consequent. 

4. An Indirect Ratio is the quotient of the consequent 
by the antecedent. 

5. A ratio of Greater Inequality is a ratio greater than 
unity; as, 7:3. 

6. A ratio of Less Inequality is a ratio less than unity; 
as, 4 :5. 

7. A Compound Ratio is the product of the correspond- 
ing terms of several simple ratios. Thus, the compound ratio of 
1:3, 5:4, and 7:2 is 1x5X7:3X4X2. 

8. A Duplicate Ratio is the ratio of the squares of two 
numbers. 

9. A Triplicate Ratio is the ratio of the cubes of two 
numbers ; as, a* : b*. 

10. A SubdupUcate Ha-tio is the ratio of the square 
roots of two numbers; as, \/~^: \/b. 

11. A SubtripHcate Ratio is the ratio of the cube roots 
of two numbers; as, ty'a'-tyb- 

PROPORTION. 

12. Proportion is an equality of ratios. The equality is 
indicated by the ordinary sign of equality or by the double colon, ::. 
Thus, a : b=c : d, or a : b : : c : d. 

13. The TJ.rt rentes of a proportion are the first and fourth 
terms. 

14. The JMeans are the second and third terms. 

15. A Mean Proportioned between two quantities is a 

quantity to which either of the two quantities bears the same ratio 
that the mean does to the other of the two. 

16. A Continued Proportion is a succession of equal 
ratios, in which each consequent is the antecedent of the next 
ratio. 

17. A Compound Proportion is an expression of 
equality between a compound and a simple ratio. 



124 FINKEL'S SOLUTION BOOK. 

18. A Conjoined Proportion is a proportion which 
has each antecedent of a compound ratio equal in value to its 
consequent. The first of each pair of equivalent terms is an an- 
tecedent; and the term following, a consequent. This is also 
called the "Chain rule." 

I. What is the ratio of -J- to f? 
i-r-f=i.Xf = f, the ratio, 

I. What is the ratio of 10 bu. to If bu. ? 

lObu.-r-lf bu. = 10 X T 7 o = 7,the ratio. 

I. What is the ratio of 25 apples to 75 boxes? 

Ans. No ratio ; for no number of times one will produce 
the other 

In a true proportion, we must always have greater : less :: 
greater : less or less : greater :: less : greater. The test for the 
truth of a proportion is that the product of the means equals the 
product of the extremes. 

T. If a 5-cent loaf weighs 7oz. when flour is $8 per barrel, 
how much should it weigh when flour is $7.50 per barrel? 

It should evidently weigh more. 

.-. less : greater : : less : greater. 
$7.50 : $8.00 : : 7oz : (? = 7-ftoz.) 

I. If a staff 3 feet long, casts a shadow 2 feet, how high is 
the steeple whose shadow at the same time is 75 feet? 

Since the steeple casts a longer shadow than the staff, it is evi- 
dently higher than the staff. 

.-. less : greater : : less : greater. 

2 feet : 75 feet :: 3 feet : ( ?=112ffeet.) 

I. What number is that which being divided by one more 
than itself, gives \ for a quotient? 

r l. Let -|=number. Then 

2. jxj==f or f :f + 1 :: 1 : 7, whence 

g3. 7(f )=l(f+l) or 
4. y=|~(-l; whence 

5. y=i, 

6- I=tV> and 

7. |=2 times T 1 2=^=number. 

III. .*. i=the number. 

I. What number divided by 3 more than itself gives f for 
a quotient? 



PROPORTION. 



125 



IIJ 



Let |-=the number. Then 

2 

J or, putting this in the form of a proportion, 



1+3 

2 • S 

"2" 



H-3 : : 7 : 9. 



[the product of the extremes. 



1 8 

2 

i= 
2_ 



1 ^= 1 ^— |— 21, the product of the means being equal to 



4=21 



:the number. 



III. 
I. 



II. 



±of 21=5i, and 
2 times 5i=10£= 
.-. 10^=the number. 

If 7 lb. of coffee is equal in value to 5 lb. of tea, and 3 lb. 
of tea to 13 lb. of sugar, 39 lb. of sugar to 24 lb. of rice, 
12 lb. of rice to 7 lb. of butter, 8 lb. of butter to 12 lb. 
of cheese ; how many lb. of coffee are equal in value to 
65 lb. of cheese? 
'1. 7 lb. of coffee=5 lb. of tea, 

2. 3 lb. of tea=13 lb. of sugar, 

3. 39 lb. of sugar=24 lb. of rice, 

4. 12 lb. of rice=7 lb. of butter, 

5. 8 lb. of butter=12 lb. of cheese, and 

6. 65 lb. of cheese— r=39 lb. of coffee, 
7X3X39X12X8X65 

'" 5X13X24X7X12 
III. ' .-. 65 lb. of cheese=39 lb. of coffee. 

I. I can keep 10 horses or 15 cows on my farm ; how many 
horses can I keep if I have 9 cows? 
15 cows : 9 cows : : 10 horses : ?=6 horses. 
10 horses — 6 horscs=4 horses. 
.-. I can keep 4 horses with the 9 cows. 
I. If 2 oxen or 3 cows eat one ton of hay in 60 days, how 
long will it last 4 oxen and 5 cows? 

2 oxen I 4 oxen : : 3 cows : ?=6 cows. 
.-. 4 oxen eat as much as 6 cows. If a ton of hay last 3 cows 
60 days, it will last 6 cows, which are equal to 4 oxen, and 5 
cows, or 11 cows, not so long. 

.-. 11 cows : 3 cows : : 60 days : ?=17 T 3 T days. 

I. If 24 men, by working 8 hours a day, can, in 18 days, dig 
a ditch 95 rods long, 12 feet wide at the top, 10 feet wide at the 
bottom, and 9 feet deep; how man}- men, in 24 days of 12 hours a 
day, will be required to dig a ditch 3S0 rods long, 9 feet wide at 
the top, 5 feet wide at the bottom, and 6 feet deep? 
95 rods : 380 rods 



24 days : 18 days 
12 hours : 8 hours 
12 feet : 9 feet 
10 feet : 5 feet 
9 feet : 6 feet 



24 men : ?=12 men. 



126 FINKEL'S SOLUTION BOOK. 



II. 



380X18 X 8 X9X5X6X24 
95X24X12X12X10X9 



=12 men. 



A Louisville merchant wishes to pay $10000, which he 
owes in Berlin. He can buy a bill of exchange in Louis- 
ville on Berlin at the rate of $.96 for 4 reichmarks ; or 
he is offered a circular bill through London and Paris, 
brokerage -J% at each place, at the following rates : 
JE'1=$4. 90=25.38 francs, and 5 francs=4 reichmarks. 
What does he gain by direct exchange? 

1. $.238=1 mark. 

2. $10000=10000-^.238=42016.807 marks. 

3. $.24=1 mark, since this is the rate of exchange. 

4. /. $10084.033=42016-807 times $.24=42016.807 marks 
=direct exchange. 

5. 42016.807 marks=( ?=$10165.38.) 

6. $4.90=£1— 1% of £l=£.99f 

7. JCl— 99£ times 25.38 fr. 

8. 5 fr.=4 marks. 

Q 42016.807X4.90X5 4:in1 , r OQ fl • L chan g e - 

9- tttt^ t^it- ^r- oo — -,=$10165. 38=cost by circular ex- 

•99|X99|X25.38X4 J 

10. $10165.38— $10084.033=$81.35 = gain by direct ex- 
change. 

III. .-. $81.35=gain by direct exchange. 

I. A wheel has 35 cogs; a smaller wheel working in it, 26 cogs; 
in how many revolutions of the larger wheel will the 
smaller one gain 10 revolutions? 

1. 35 cogs — 26 cogs=9 cogs=what the smaller wheel gains 
on larger in 1 revolution of larger wheel. 

2. 26 cogs passed through the point of contact=l revolu- 
tion of smaller wheel. 

3. 1 cog passed through the point of contact=^ revolu- 
tion of smaller wheel. 

4. 9 cogs passed through the point of contact=^g revolu- 
tion of smaller wheel. 

5. .". In 1 revolution of larger wheel the smaller gains ^ 
revolution of smaller wheel. 

6. .*. -9 9 g- revolution gained '. 10 revolutions gained : : 1 
revolution of larger wheel : ?=28f revolutions of 
larger wheel. 

III. .*. The smaller wheel will gain 10 revolutions in 28-f revo- 
lutions of larger wheel. 

By analysis and proportion. 

26 cogs passed through the point of contact=l revolution of 
the smaller wheel. 



II. 



PROBLEMS. 127 

35 cogs passed through the point of contact=l revolution of 
the larger wheel. But when the larger wheel has made 1 revo- 
lution, 35 cogs of the smaller wheel have passed through the 
point of contact. If 26 cogs having passed through the point of 
contact make 1 revolution of the smaller wheel, how many rev- 
olutions will 35 cogs make? 

By proportion, 26 cogs : 35 cogs : : 1 rev. : ?=1 ^ rev. 

.-. The smaller wheel makes l-^g- revolutions while the larger 
wheel makes 1 revolution. .*. The smaller gains 1^ revolutions 
— 1 revolution=^g- revolution. If the smaller wheel gains -^ 
revoluion in 1 revolution of the larger wheel to gain 10 revolu- 
tions on the larger wheel, the larger wheel must make more rev- 
olutions.- .*. less : greater : : less I greater. 

-^g- rev. ; 10 rev. ; I 1 rev. of larger ; ?=28f rev. of larger. 

I. If the velocity of sound be 1142 feet per second, and the 
number of pulsations in a person 70 per minute, what 
is the distance of a cloud, if 20 pulsations are counted 
between the time of seeing the flash and hearing the 
thunder? 

1. 1142 ft.=distance sound travels in 1 second. 

2. 68520 ft.=60Xll42 ft.=distance sound travels in 1 min., 
or the time of 70 pulsations. 

3. .'. If it travels 68520 feet while 70 pulsations are count- 
ed, it will travel not so far while 20 pulsations are 
counted. 

4. .*. greater : less : : greater : less. [145 yd. 2y ft. 

5. 70 pul. : 20pul. : : 68520 ft. : ?=19577| ft.=3 mi. 5 fur. 

III. .-. The cloud is 3 mi. 5 fur. 145 yd. 2\ ft. distant. 

(/?., 3d /., p. 2'89, prob. 45.) 

PROBLEMS. 

1. If 3 horses, in £ of a month eat f of a ton of hay, how long 
will f of a ton last 5 horses? 

2. If a 4-cent loaf weighs 9 oz. when flour is $6 a barrel, how 
much ought a 5-cent loaf wei«h when flour is $8 per barrel? 

3. A dog is chasing a hare, which is 46 rods ahead of the dog. 
The dog runs 19 rods while the hare runs 17; how far must the 
dog run before he catches the hare? 

4. If 52 men can dig a trench 355 feet long, 60 feet wide, and 
8 feet deep in 15 days, how long will a trench be that is 45 feet 
wide and 10 feet deep, which 45 men can dig in 25 days? 

5. If \ of 12 be 3 what will J of 40 be? Ans. 15. 

6. If 3 be l of 12, what will £ of 40 be? Ans. 6|. 



II. 



128 FINKEL'S SOLUTION BOOK. 

7. If 18 men or 20 women do a work in 9 days, in what time 
can 4 men and 9 women do the same work? Ans. 62V days. 

8. If 5 oxen or 7 cows eat 3 T 4 T tons of hay in 87 days, in what 
time will 2 oxen and 3 cows eat the same quantity of hay? 

Ans. 105 days. 

9. Divide $600 between three men, so that the second man 
shall receive one-third more than the first, and the third § more 
than the second. 

10. Two men in Boston hire a carriage for $25, to go to Con- 
cord, N. EL, and back, the distance being 72 miles, with the 
privilege of taking in three more persons. Having gone 20 
miles, they took in A ; at Concord they took in B ; and when 
within 80 miles of Boston, they took in C. How much shall 
each pay? Ans. First man, $7.609j£f ; second, $7.609f£§ ; A, 
$5.873 T y ¥ 5 1^ $2,8G4 T V ; and C, $1.041 T V 

11. Three men purchased 6150 sheep. The number of A's 
sheep is to the number of B's sheep as § is to 3£, and 4 times the 
number of C's sheep is to the number of A's sheep as ^ is to \. 
Find the number of sheep each had. C A's= 

Ans. ] B's = 
( C's = 

12. If $500 gain $10 in 4 months, what is the rate per cent? 

Ans. 8%. 

13. If 12 men can do as much work as 25 women, and 5 wo- 
men do as much as 6 boys ; how many men would it take to do 
the work of 75 boys? Ans. 30 men. 

14. If 5 experienced compositors in 16 days, 11 hours each, 
can compose 25 sheets of 24 pages in each sheet, 44 lines on a 
page, 8 words in a line, and 5 letters to a wofd ; how many in- 
experienced compositors in 12 days, 10 hours each, will it take 
to compose a volume (to be printed with the same kind of type), 
consisting of 36 sheets, 16 pages to a sheet, 100 lines to the 
page, 5 words to a line, and 9 letters to a word, provided that 
while composing an inexperienced compositor can do only |- as 
much as an experienced compositor, and that the latter work is 
only I as hard as the former? Ans. 16. 

15. If A can do f as much in a day as B, B can do fas much 
as C, and C can do -J as much as D, and D can do f as much as 
E, and E can do ^ as much as F; in what time can F do as much 
work as A can do in 28 days? Ans. 8. 

16. A starts on a journey, and travels 27 miles a day; 7 days 
after, B starts, and travels the same road, 36 miles a day; in how 
many days will B overtake A? Ans. 21 days. 



ANALYSIS. 129 

17- A wheel has 45 cogs ; a smaller wheel working in it, 36 
cogs ; in how many revolutions of the larger wheel will the 
smaller gain 10 revolutions? Ans. 40. 

18. If the velocity of sound be 1142 feet per second, and the 
number of pulsations in a person 70 per minute, what is the dis- 
tance of a cloud, if 30 pulsations are counted between the time of 
seeing a flash of lightning and hearing the thunder? 

Ans. 5£ mi. 108 yd. If ft. 

19. If William's services are worth $15f a month, when he 
labors 9 hours a day, what ought he to receive for 4f months, 
when he labors 12 hours a day? Ans $91.91^. 

20. If 300 cats kill 300 rats in 300 minutes, how many cats 
will kill 100 rats in 100 minutes? Ans. 300 cats. 



CHAPTER XVII, 

ANALYSIS. 



1. Analysis, in mathematics, is the process of solving 
problems by tracing the relation of the parts. 

I. What will 7 lb. of sugar cost at 5 cents a pound? 

Analysis for primary classes. 

If one ponnd of sugar costs 5 cents, 7 pounds will cost 7 times 
5 cents, which are 35 cents. 

I. If G lead pencils cost 30 cents, what will one lead pencil 
cost? 

Analysis: If 6 lead pencils cost 30 cents, one lead pencil will 
cost as many cents as 6 is contained into 30 cents which are 5 
cents. 

I. If 8 oranges cost 48 cents, what will 5 oranges cost? 

Analysis: If 8 oranges cost 48 cents, one orange will cost as 
many cents as 8 is contained into 48 cents which are 6 cents; if 
one orange costs 6 cents 5 oranges will cost 5 times G cents, which 
are 30 cents. 

I. If a boy had 7 apples and ate 2 of them, how many had he 
left? 

Analysis: If a boy had 7 apples and ate 2 of them, he had 
left the difference between 7 apples and 2 apples which are 5 
apples. 

I. If John had 12 cents and found 5 cents, how many cents 
did he then have? 

Analysis: If John had 12 cents and found 5 cents, he then 
had the sum of 12 cents and 5 cents which are 17 cents. 



130 



FINKEL'S SOLUTION BOOK. 



Note. — If teachers in the Primary Departments would see that their 
pupils gave the correct analysis to such problems, their pupils would often 
be better prepared for the higher grades. After they are thoroughly ac- 
quainted with the analysis of such questions they may be taught to write 
out neat, accurate solutions with far less trouble than if allowed to give 
careless analysis to problems in the lower grades. 

I. If 4 balls cost 36 cents, how many balls can be bought fur 
81 cents? 

Analysis: If 4 balls cost 36 cents, one ball will cost as many 
cents as 4 is contained into 36 cents which are 9 cents; if one ball 
costs 9 cents for 81 cents there can be bought as many balls as 9 
is contained into 81 which are 9 balls. 

Written solution. 

1. 36 cents=cost of 4 balls. 
II. \ 2. 9 cents=36 cents-T-4=cost of 1 ball. 
3. 81 cents=cost of 81-r-9, or 9 balls. 

III. .-. If 4 balls cost 36 cents, for 81 cents there can be bought 
9 balls. 



I. What number divided by | will give 10 for a quotient? 



II.< 

III. 
I. 

II.< 
III. 



1. |=the number. 



f-l==f X 4=f=quotient 



\ of 10=2, and 
— 3 times 2=6=the number. 



5 • 3 

5 • 5 

3. 10=quotient. 

4. ..-. f=10, 

5. i= 

6. | 

.-. 6= the number required. 

$25 is J of the cost of a barrel of wine; what did it cost? 

1. |-=cost of the wine per barrel. 

2. | of cost=$24, 

3. i of cost=i of $24=48, 
.4. | of cost=5 times $8=4 

.-. $40=cost of wine. 



What number is that from which, if you take -^ of itself, 
the remainder will be 16? 



IU 



III. 



1. ^=the number. 

2. j — |-=^=remainder after taking away |-. 

3. 16=remainder. 

4. .-. f=16, 

5. f=iof 16— 4, and 

6. y=7 times 4=28=the number. 

.-. 28==the required number, 



ANALYSIS. 



131 



I. 



II.< 



III. 



A boat is worth $900; a merchant owns f of it, and sells^ 
of his share ; what part has he left, and what is it 
worth? 



■{ 



B. 



rl. 
J 2- 

r 



f=part the merchant owned. 

4- of f=2 5 ¥ =part he sold. 

.-. I— A=M— 2 5 4=M-fV=Part he had 

$900=value of if, or the whole ship. 

$75= T V of $900=value of ^ of the ship 



left. 



$375=5 
he had 



times $75=value of -^ of the ship, or part 



left. 



y 5 2=part he had left, and 
$375=value of it. 



II. 



A and B were playing cards. B lost $14, which was T 7 ^ 
times § as much as A then had ; and when they com- 
menced, | of A's money equaled -f- of B's. How much 
had each when they began to play? 



-ft of B's. 



| of A's money=4 of B's. 

^ of A's money=i of -|= 

| of A's money=8 times ^=^4 of B's. 

|§=B's money when they began to play. Then 

^f=A's money when they began. 

if = A's money after winning $14 from B. 

$14=what B lost. 

T 7 ^ times !-== T 7 -j==part A's money is of $14. 

• •• tV=$14, 

^=4 of $14=$2, and [$14 from B. 

-j-|=15 times $2=$30=A's money after winning 

.-. $30— $14=$ 16= A's money at first. 

.-. H=$16, from (5), 

^= T Vof$16=$l,and 

§-§=35 times $l=$35=B's money at first. 

$16=A's money at first, and 
$35=B's money at first. 

(Stod. Int. A.,p. lll,p'ob. 30.) 

I. A drover being asked how many sheep he had, said, if to 



(1.) 

(2.) 
(3.) 
(4.) 
(5.) 



(6.) 



(V.) 

(8.) 

(9.) 

(10.) 



n. 

2. 
3. 
4. 
5. 
6. 



III. 



-£ of my flock you add the number 
99^-; how many sheep had he? 

1. |=the number of sheep. 

2. |-f-9|=i of the number+9f 



the sum will be 



II. 



III. 



99^ 



:-£ of the number+9^. 
... J+9^99^ or 
i=99|— 9|=90, and 
|=3 times 90=270=number of sheep. 

He had 270 sheep. 



132 



FINKEL'S SOLUTION BOOK. 



I. 



II. 



III. 



Heman has 6 books more than Handford, and both have 
26; how many have each? 

|-=number Handford has. Then 
|— |-6=He man's number. 



f-f-f + 6=1+6— number both have. 

26=number both have. 

.-. 4+6=26 or 

f=26—6=20. 

^=1- of 20=5, and 

f=2 times 5=10=Handford's number. 

|+6=16=Heman's number. 

Handford had 10 books, and 

Heman had 16 books. (Stod. Int. A., p. 116,prob. 2.) 



A man and his wife can drink a keg of wine in 6 days, 
and the man alone in 10 days ; how many days will it 
last the woman ? 



II. 



III. 



a 

2. 

3. 
4. 
5. 

6. 



6 days=time it takes both to drink it. 
l=part they drink in one day. 
10 days=time it takes the man to drink it. 
T 1 Q-=pait he drinks in one day. 



[day. 
one 



.-. I — T L==-^y — ¥ 3 ^= T 1 - 3 -=part the woman drinks in 
l-|=what the woman drinks in y|~— ^=15 days. 

It will take the woman 15 days. 

(R. Alg. I., p. 112, prod. 59.) 



I. A man was hired for 80 days, on this condition: that for 
every day he worked he should receive 60 cents, and 
for every day he was idle he should forfeit 40 cents. At 
the expiration of the time, he received $40. How many 
days did he work? 

1. $.60=what he receives a day. 

2. $48=80 X$.60=w hat he would have leceived had he 
worked the whole time. 

3. $40=what he received. 
II.<J4. .-. $48 — $40=$8=what he lost by his idleness. 

5. $1=$.60, his wages,+$.40, what he had to forfeit,= 
what he lost a day. 

6. .'. $8=what he lost in 8-—1, or 8 days. 
17. 80 days — 8 days=72 days, the time he worked. 

III. .-. He worked 72 days. 

I. A ship-mast 51. feet high, was broken off in a storm, and 
§ of the length broken off, equaled f of the length re- 
maining; how much was broken off, and how much re- 
mained ? 






ANALYSIS. 133 

1. f of length broken ofT=j of length remaining, 

2. ■§• of length broken off=^- of f=-§ of length remaining, 

3. f of length broken off=3 times ■§=§ of length remain- 
ing. 

4. |=length remaining. 
i 5. |=length broken off. 

6. f+|=Y=whole length. 

7. 51 feet=whole length. 

8. .'. V=51 feet, 

9. i= T \ of 51 feet==3 feet, and 

10. f=8 times 3 feet=24 feet, length remaining. 

11. |=9 times 3 feet=27 feet, length broken off. 

Uj J 24 feet=length remaining, and 

'"• I 27 feet=length broken off. 

I. A boy being asked his age, said, "4 times my age is 24 
years more than 2 times my age;" how old was he? 

1. |=his age. 

2. 4xf=~|=4 times his age. 

3. 2Xf=|=2 times his age. 
II.<|4. ... ! = 4_|_24 years or 

5. | — 4 = 4 = 24 years. 

6. |=i of 24 years=6 years, and 

7. J=2 times 6 years=12 years, his age. 

III. .'. He is 12 years old. (Stod. Int. A., p. 116, prod. 16.) 

I, If 10 men or 18 boys can dig 1 acre in 11 days, find the 
number of boys whose assistance will enable 5 men to 
dig 6 acres in 6 days. 

1. 1 A.=what 10 men dig in 11 days. 

2. jq A.=what 1 man digs in 11 days. 

3. jyo A.= TT of y 1 ^ A.=what 1 man digs in 1 day. 

4. yV A.=yf-ff A.=5 times TT7 A.=what 5 men dig in 1 
day. [days. 

5. T 3 T A. =2% A. =6 times ■£% A.=what 5 men dig in 6 

6. .*. 6 A. — T ; T A.^Sy 8 ! A.=what is to be dug by the boys 
in 6 days. 

7. 1 A.=what 18 boys dig in 11 days. 

8. y 1 ^ A.=what 1 boy digs in 11 days. 

9. T ^g A.= TT of Jg- A.=what 1 boy digs in 1 day. 

10. -gV A.= T f^ A.=G times T ^ A. = what 1 boy digs in 6 
days. 

11. 5 T 8 y A.=what 5 T 8 y-:- ^ , or 189, boys dig in 6 days. 

III. .-. It will take 198 boys. 

(i?. 3d p., O. B.,p. SIS, prod. 66.) 



134 FINKEL'S SOLUTION BOOK. 

I. A man after doing f of a piece of work in 30 days, calls 
an assistant; both together complete it in 6 days. In 
what time could the assistant complete it alone? 

1. f=part the man does in 30 days. 

2. ^L=Jq- of f=part he does in 1 day. 

3. f=f — f=P ar t he and the assistant do in 6 days. 
II.<l4. T X 5=i of -|— part he and the assistant do in 1 day. 

5 - •'• tV-A==ttV-- TTo=Tir o=P art the assistant does in 1 

day. 
6. if^=part the assistant does in Tii~^~ii'U == ^7 d a y s - 

III. .*. It will take the assistant 21f days. 

(R. 3d p., O. E.,p. 318, prod. 71.) 

Explanation. — Since the man does § of the work before he called on the 
assistant, there remains § — |=fj which he and the assistant do in 6 da^ s. 
Hence they do £ off, or j^g of the work in one day. If the man and his 
assistant do ^ of the work in 1 day and the man does ^ of the work in 1 
day, the assistant does the difference between^ and 5 J 5 which is r f of the 
work in 1 day. Hence it will take ^fg-j-j^Q, or 21f days, to do the work. 

I. A person being asked the time of day, replied that it was 
past noon, and that f of the time past noon was equal 
to f of the time to midnight. What was the time of 
day? 

1. ■§ of the time past noon=>| of the time to midnight. 

2. \ of the time past noon— ^ of f=^ of the time to mid- 
night, [midnight. 

3. f, or the time past noon,=4 times \=^ of the time to 

4. |=time to midnight. Then 
\\.{ 5. ^=time past noon. 

6. ^-|— !=!=time from noon to midnight. 

7. 12 hours =time from noon to midnight. 

8. .'. |=12 hours, 

9. \=\ of 12 hours=l^ hours, and [past noon. 
10. f=4 times 1^ hours=5^ hours=5 hr. 20 min., time 

III. .-. It is 20 min. past 5 o'clock, P. M. 

[Milne's Prac. A., p. 360, prod. 1^7.) 

Note. — From 3, we have the statement that the time past noon is § of the 
time to midnight. Hence, if § is the time to midnight, f is the time past 
noon or if \% is the time to midnight, r 8 5 is the time past noon. 

I. A person being asked the time of day, said that f of the 
time past noon equals the time to midnight. What is 
the time of day ? 



ANALYSIS. 1$5 



II. 



'1. ^-=time past noon. Then 

2. ^=time to midnight. 

3. s.-J-T.— l^—time from noon to midnight. 

4. 12 hours=time from noon to midnight. 



5. .'. 1 T 2 =12 hours. 



-- - - i -— ■ 

6. \=-T2 of 12 hours=l hour, and 

7. y=7 times 1 hour=7 hours=time past noon. 

III. .-. It is 7 o'clock P. M. 

I. A man being asked the hour of day, replied that ^ of the 
time past 3 o'clock equaled ^ of the time to midnight; 
what was the hour? 

1. ^ of the time past 3 o'clock=-| of the time to midnight. 

2. \ , or the time past 3 o'clock,=4 times i=f of the time 
to midnight. 

3. |-=time to midnight. 

4. J=time past 3 o'clock. 
II.<! 5. |-|-f=f=time from 3 o'clock to midnight. 

6. 9 hours=time from 3 o'clock to midnight. 

7. .-. f=9 hours. 

8. \=\ of 9 hours=l^ hours, and 

9. f= 4 times 1-^ hours=6 hours=time past 3 o'clock. 
10. |— 1-3 hours=9 hours, time past noon. 

III. .-. It is 9 o'clock, P. M. 

(Brooks' Int. A.,f. 156,prob. 17.) 

I. A person being asked the hour of day, replied, § of the 
time past noon equals f of time from now to midnight 
-f-2f- hours; what was the time? 

1. f of time past noon=f of time to midnight-(-2§ hours. 

2. -J of time past noon=-J of (f-f-2f hours)=i of time to 
midnight+1^ hours. [to midnight-j-4 hours. 

3. f, or time past noon,=3 times (-§-+!-£ hours)=^ of time 

4. f=time to midnight. 
tt ) 5. l-f-4 hours=time past noon. [night. 

6. | | \ [ 4 hours=f-|-4 hours=time from noon to mid- 

7. 12 hours=time from noon to midnight. 

8. .-. 1+4 hours=12 hours. 

9. 4=12 hours — 4 hours=8 hours, 

10. \=\ of 8 hours=2 hours, and 

11. J-f-4 hours=6 hours=time past noon. 

III. .-. It is 6 o'clock, P. M. 

(Stod. Int. A., p. 128, Prob. 29.) 

I. A father gave to each of his sons $5 and had $30 remain- 
ing; had he given them $8 each, it would have taken all 
his money; required the number of sons. 



II. 



136 FINKEL'S SOLUTION BOOK. 

'1. $8=amount each received by the second condition. 
2. $5=amount each received by the first condition. 
II.<J 3. $3=$8 — $5= excess of second condition over first, on 
each son. [10 sons. 

.4. .*. $30=excess of second condition over first, on 30-r-3, or 

III. .*. There were 10 sons. 

I. If 50 lb. of sea water contain 2 lb. of salt, how much fresh 
water must be added to the 50 lb. so that 10 lb. of the 
new mixture may contain -J lb. of salt. 

1. -^ lb. of salt=what 10 lb. of the new mixture contains. 

2. f, or 1, lb. of salt=what 3 times 10 lb., or 30 lb., of the 
new mixture contain. [mixture contain. 

3. 2 lb. of salt=what 2 times 30 lb., or 60 lb., of the new 

4. .-. 60 lb.— 50 lb.=10 lb.=quantity of fresh water that 

must be added. 

III. .-. 10 lb. of fresh water must be added that 10 lb. of the 
new mixture may contain -^ lb. of salt. 

I. A farmer had his sheep in three fields, § of the number 
in the first field equals f of the number in the second 
field, and § of the number in the second field equals f 
of the number in the third field. If the entire num- 
ber was 434, how many were in each field? 

1. -| of number in first field=f of number in second 

field. [second field. 

(l.)<!2. -J of number in first field=^ of f=§ of number in 

|^, or number in first field,=3 times |=-f of number 

in second field. 

II. J of number in second field=f of number in third 
field. [in third neld. 

2. -J of number in second fieid=-§- of f=f of number 
3. f, or number in second field,=3 times -§— f of num- 
ber in third field. 
(3.) -|=number in third field. Then 
(4.) -|=number in second field, and 
IlJ (5.) -|i=|- of number in second field=number in first 
field in terms of number in third field. 
(6.) .•■f + * + ti=H+H+H=W=number in the 

three fields. 
(7.) 434=number in the three fields. 
(8.) .*. W=434, 

(9.) ^=2^ of 434=2, and [field. 

(10.) f|=64 times 2=128=number of sheep in third 

(11.) -g-f=72 times 2=144=number of sheep in second 

field. [field. 

.(12.) -|^=81 times 2=162=number of sheep in first 



ANALYSIS. 137 

( 162=number of sheep in first field, 
III. .-. < 144=number of sheep in second field, and 
( 128=number of sheep in third field. 

{Milne's Prac. A., p. 362, f rob. 68.) 

I. In a certain school of 80 pupils there are 32 girls ; how 
many boys must leave that there may be 5 boys to 4 
girls? 

1. 80=whole number of pupils. 

2. 32=number of girls. 

3. 80— 32=48=number of boys. 

4. |=number of girls. Then, since the number of boys are 
to be to the number of girls as 5 : 4, 

II.<j5. |=number of boys. But 

6. - 32. 

7. i=i of 32=8, and 

8. f=5 times 8=40=number of boys. 

9. .". 48 — 40=8=number that must leave that there may be 
5 boys to 4 girls. 

III. .'. 8 bo\s must leave that there may be 5 boys to 4 girls. 

I. How far may a person ride in a coach, going at the rate 
of 9 miles per hour, provided he is gone only 10 hours, 
and walks back at the rate of 6 miles per hour? 

1. 9 mi.=distance he can ride in 1 hour. 

2. 1 mi.=distance he can ride in \ hour. 

3. 6 mi.=distance he can walk in 1 hour. 
II.<|4. 1 mi.=distance he can walk in \ hour. 

5. •'. \ hr.-f-i hr.= T 5 g hr.=time it takes him to ride 1 mi. 
and walk back. [and walk back. 

6. .'. 10 hours=time it takes him to ride 10-H- T 5 ¥ , or 36, mi. 

III. .'. He can ride 36 miles. 

I. A hound ran 60 rods before he caught the fox, and f- of 
the distance the fox ran before he was caught, equaled 
the distance he was ahead when they started. How far 
did the fox run, and how far in advance of the hound 
was he when the chase commenced? 

1. |=distance the fox ran before he was caught. Then 

2. f=distance he was ahead. 

3. |-|-§=-§=di stance the hound ran to catch the fox. 

4. 60 rods=distance the hound ran to catch the fox. 
II.<!5. .-. f=60 rods, 

6. \=\ of 60 rods=12 rods, and [ahead. 

7. f=2 times 12 rods=24 rods=distance the fox was 

8. |=3 times 12 rods=36 rods=distance the fox ran be- 
fore he was caught. 



133 FINKEL'S SOLUTION BOOK. 

TJT ( 24 rods=distance the fox -was ahead, and 

I 36 rods=distance he ran before he was caught. 

I. If i of 12 be 3 , what will I of 40 be ? 

1. i of 12=4. 

2. £ of 40=10. By supposition 
IL<j3. 4=3. Then 

4. 1=J of 3=f, and 
15. 10=10 times j=7i- 

III. .*. J of 40=7-J, on the supposition that ^ of 12 is 3. 

I. Eight men hire a coach; by getting 6 more passengers, 
the expenses of each were diminished $lf ; what do they 
pay for the coach? 

1. -|=amount paid tor the coach. [been only 8 men. 

2. -J=amount 1 man would have had to pay, had there 

3. T 1 ¥ =amount 1 man paid since there were 8 men-)-6 men, 
or 14 men. 

Il.H. .-. -J— T i ¥ = T V— A=A =what each sa ved. 

5. $lf=what each saved. 

6. .-. *=*!*. 

7- -h=i of m=^r» and 

.8. ||=56 times $ T 7 ^=$32f=amount paid for the coach. 

III. .*. $32f=amount paid for the coach. 

(R. H. A.,p. 40S,prob. 46.) 

Second solution. 

(1. $lf=amount saved by each man. [the six men. 

TT J 2. $14=8X$11— amoun t saved by the 8 men and paid by 

'1 3. .'. $2^=^ of $14=amount paid by each of the 14 men. 

U. .-. $32f=14 times $2^=amount they paid for the coach. 

III. .-. They paid $32f for the coach. 

I. For every 10 sheep I keep I plow an acre of land, and 
allow one acre of pasture for every 4 sheep; how many 
sheep can I keep on 161 acres? 

1. 1 A.=what I plow for every 10 sheep I keep. 

2. T \A.=what I plow for each sheep I keep. 

3. lA.=what I allow for pasture for every 4 sheep I keep. 

4. ¥ A.=what I allow for pasture for each sheep I keep. 

5. .-. 1 i QA.-)-^A.==^ -A.=land required for every sheep. 

6. .-. 161A.=land required for 161-7-^, or 460 sheep. 

III. .*. I can keep 460 sheep on 161 acres. 

(R.Alg. I.,p.ll2,prob.64.) 
Complete analysis. 

If for every 10 sheep I plow 1 acre, for 1 sheep I plow ^ of 
an acre ; and if for every 4 sheep I pasture 1 acre, for 1 sheep, I 



II. 



ANALYSIS. 139 

pasture \ of an acre ; hence 1 sheep requires ^A.-f-JA., or ^A., 
and on 161 A. I could keep as many sheep as ^g-A. is contained in 
161 A., which are 460 sheep. 



A man was engaged for one year at $80 and a suit of 
clothes; he served 7 months, and received for his wages 
the clothes and $35; what was the value of the clothes? 

1. -i-|=value of the suit of clothes. 

2. j-f+$ 80= wages for 1 year or 12 months. 

3. T 1 2+$ 6 t=T 1 2 of (if+$80)=wages for 1 month. 

4. T 7 2-)-$46f=7 times (yV+'^t )=wages for 7 months. 
II. ) 5. ||_j_^35 =wa g es f or 7 months. 



7. - 

8. J"=4 of $llf=$2£, and 
19. lf=12 times $2£=$28=value of suit of clothes. 

III. .-. The suit of clothes is worth $28. 



I. A lady has two silver cups, and only one cover. The 
first cup weighs 8 ounces. The first cup and cover 
weighs 3 times as much as the second cup; and the sec- 
ond cup and cover 4 times as much as the first cup. 
What is the weight of the second cup and the cover? 



1. 3 times weight of second cup==weight of cover-j-weight 
of first cup, or 8 oz. [2f oz. 

2. 1 times weight of second cup=^- of weight of cover-|- 

3. §=weight of cover. Then 

4. |~(-2§ oz.=weight of second cup. [cover. 

5. |+i+2f oz.=f+2f oz. = weight of second cup and 
II. I 6. 32 oz.=4 times 8 oz.=weight of second cup and cover, 

by the conditions of the problem. 

7. .-. |+2f oz.=32 oz. 

8. |=32 oz.— 2f oz.=294 oz - 

9. |=4 of 29£ oz.=7£ oz. 

10. f=3 times 7^ oz.=22 oz.=weight of cover. [cup. 

11. -J+2f oz.=7i oz.-(-2f oz.= 10 oz. = weight of second 

TTT . { 22 oz.=weight of cover, and 
( 10 oz.=weight of second cup. 



A steamboat that can run 15 mi. per hr. with the current 
and 10 mi. per hr. against it, requires 25 hr. to go from 
Cincinnati to Louisville and return ; what is the dis- 
tance between the cities? 



140 FINKEL'S SOLUTION BOOK. 



15 mi.=distance the boat can travel down stream in 
1 hour. [hour. 

*« ft 



2. 1 mi.=distance the boat can travel down stream 



3. 10 mi.=distance the boat can travel up stream in 1 hr, 

II. ^4. 1 mi.=distance the boat can travel up stream in y 1 -^ hr. 

.-. y 1 ^ hr.-j-y 1 ^ hr.=i hr.=time required for the boat to 

travel 1 mi. down and return. 
.". 25 br.=time required for the boat to travel 25-f-J, or 
150, miles down and return. 

III. .*. The distance between the two places is 150 miles. 

I. A, B, and C dine on 8 loaves of bread ; A furnishes 5 
loaves ; B, 3 loaves; C pays the others 8d. for his share; 
how must A and B divide the money? 

1. 8 loaves=what they all eat. 

2. 2-| loaves=what each eats. 

3. .". 5 loaves — 2f loaves=2^ loaves=what A furnished 
towards C's dinner. 

4. .*. 3 loaves — 2f loaves=^ loaf=what B furnished to- 
wards C's dinner. 



II. 



2 j 

* J ^— =^=A's share, and 

'2 o 



(1, 



21 

1 



K2— -J=B's share. 

7. £ of 8d.=7d.=what A should receive, and 
.8. i of 8d.=ld.=what B should receive. 



8 



TTT -. { A should receive 7d., and 

" •'"". ( B should receive Id. (A*. H. A., />. £03, prob. ^2.) 

I. A and B dig a ditch 100 rods long for $100; how many 
rods does each dig, if they each receive $50, and A digs at $.75 
per rod, and B at $1.25? 

There has been a vast amount of quibbling about this problem; 
but a few moments consideration should suffice to settle all dis- 
pute, and pronounce upon it the sentence of absurdity. 

We have given, the whole amount each received and the 
amount each received per rod. Hence, if we divide the whole 
amount each received by the cost per rod, it must give the num- 
ber of rods he digs. But by doing this we receive 50-f-.75, or 66f 
rods, what A digs and 50-f-1.25, or 40 rods, what B digs, or 
106f rods which is the length of the ditch, and not 100 rods as 
stated in the problem. The length of the ditch is a function of 
the cost per rod and the whole cost, and when they are given 
the length of the ditch is determined. We might propose a 
problem just as absurd by requiring the circumference of a circle 
whose area is 1 acre, and diameter 20 rods. Since the area 
and circumference are functions of the diameter, when either 



ANALYSIS. 141 

of these are given, the other is determined and should not be 
limited to an inaccurate statement. 

If, in the original problem, A's price per rod increases at a 
constant ratio so that when the ditch is completed he is receiving 
$1 per rod, and B's price constantly decreases until when the 
ditch is completed he is receiving $1 per rod, then the problem 
is solvable, and the result is 50 rods each. 

I. A is 30 years old, and B is 6 years old ; in how many 
years will A be only 4 times as old as B? 

1. |=B's age at the required time. Then 

2. -§=A's age at the required time. 

3. -f — f=f=difference of their ages. 

4. 30 years — 6 years=24 years=dirTerence of their ages. 



IT. 



5. .-. |=24 vears. 



•J=£ of 24 years=4 years. [time. 

|=2 times 4 years=8 years, B's age at the required 
.*. 8 years — 6 years=2 years=the number of years hence 
when A will be only 4 times as old as B. 

III. .-. In 2 years A will be only 4 times as old as B. 

I. Jacob is twice as old as his son who is 20 years of age ; 
how long since Jacob was 5 times as old as his sou? 

1. 20 years=son's age at present. Then 

2. 40 years— Jacob's age at present. 

3. f=son's age at required time. Then 

4. y^Jacob's age at required time. 



II. 



y — f=£=difference of their 



(>. 40 years — 20 years=20 vears=diflerence of their ages. 

7. .-. f=20 years, 

8. ^—i of 20 years=2.} years, and [ time. 

9. f=2 times 2 J years=5 years, son's age at the required 
10. .'. 20 years — 5 years=15 years=:time since Jacob was 5 

times as old as his son. 
ITT. .-. 15 years ago Jacob was 5 times as old as his son. 

Remarks. — Observe that the difference between any two persons' ages is 
constant, that is, if the difference between A's and B's ages is 7 years now, 
it will be the same in any number of 3 ears from now; for, as a year is add- 
ed to one's age, it is likewise added to the other's age. But the ratio of 
their ages is constantly changing as time goes on. If A is 3 years old and 
B 5 years old, A is now % as old as B; but in 1 year, A's age will be 4 vears 
and B's years; A is then * as old as 15. In 7 years, A will be 10 years old 
and B 12; A will then be {!,', or |, as old as P>, and so on. The ratio of any 
two persons' ages approaches unity as its limit. 

I. A fox is 50 leaps ahead of a hound, and takes I leaps in 
the same time that the hound takes 3; but 2 of the 
hound's leaps equal 3 of the fox's leaps. IIow many 
leaps must the hound take to catch the fox? 



142 



FINKEL'S SOLUTION BOOK. 



II. 



III. 



1. 2 leaps of hound's=3 leaps of fox's. 

2. 1 leap of hound's=J of 3 leaps=l-J- leaps of the fox's. 

3. 3 leaps of hound's=3 times 1^ leaps=4J leaps of fox's. 



4-J- leaps — 4 leaps=-| leap: 



hat 



II. 



III. 
I. 



the hound gains in 

taking 3 leaps. [ m g 6 leaps. 

). .*. 1 leap=2 times -^ leap=what the hound gains in tak- 

). .-. 50 leaps=what the hound gains in taking 50X6 

leaps, or 300 leaps. 
.-. The hound must take 300 leaps to catch the fox. 

Remark — We see that 3 of the hound's leaps equals A\ leaps of the fox's, 
But while the hound takes 3 leaps, the fox takes 4 leaps; hence the hound 
gains 4£ — 4, or ^, leap of the fox's. But he has 50 leaps of the fox's to gain, 
and since he gains \ leap of the fox's in 3 leaps, he must take 300 leaps to 
gain 50 leaps. 

I. If 6 sheep are worth 2 cows, and 10 cows are worth 5 
horses; how many sheep can you buy for 3 horses? 
Value of 2 cows=value of 6 sheep. 
Value of 1 cow=value of 3 sheep. 
Value of 10 cows— value of 30 sheep. But 10 cows are 

worth 5 horses, 
.*. Value of 5 horses=value of 30 sheep. 
Value of 1 horse=value of 6 sheep. 
Value of 3 horses=value of 18 sheep. 
.-. 3 horses are worth 18 sheep. 

A teacher agreed to teach a certain time upon these con- 
ditions : if he had 20 scholars he was to receive $25; 
but if he had 30 scholars, he was to receive but $30. 
He had 29 scholars. Required his wages. 
$25=his rate of wages for 20 pupils. 
$1.25=-^r of $25=his rate of wages for 1 pupil. 
$30=his rate of wages for 30 pupils. 
$l=-gL of $30=his rate of wages fcr 1 pupil. 
.-. $1.25— $1.00=$.25=reduction per pupil by the ad- 
dition of 10 pupils. 
$. 025=$. 25-r-lO— reduction per pupil by the addition of 

1 pupil. 
$.225=9 times $.025=reduction per pupil by the addi- 
tion of 9 pupils. 
.-. $1.25— $.225=$1.025=his rate of wage per pupil. 
$29,725=29 times $1.025=his wages for 29 pupils. 
His wages were $29,725. 

(Mattooris Arith., f. 385,prob. 200.) 
Note. — This problem is really indeterminate, because there is no definite 
rate of increase of wasres given for each additional scholar. We might say, 
since the wages were increased $5 by the addition of 10 scholars, they would 
be increased $.50 by the addition of one scholar and, consequently, $4.50 by 
the addition of 9 scholars. Hence, his wages should be $25-|-$4.50, or 
$29. 50. By assuming different relations between the increase of wages and 
additional scholars, other results may be obtained. The above solution 
seems to be the most satisfactory. 



II. 



III. 



1. 
2. 
3. 

4. 
5. 

6. 

7. 

8. 
,9. 



II.< 



III. 



III. 



ANALYSIS. 143 

I. A gold and silver watch were bought for $160; the silver 
watch cost only -^ as much as the gold one ; how much 
was the cost of each ? 

1. ^=cost of the gold watch. Then 

2. i=cost of the silver watch. 

3. 7_|_i = 8 ==cost f both. 

4. $160=cost of both. 

5. .'. f=$160, 

6. f=| of $160=$20=cost of the silver watch, and 

7. ^=7 times $20=$140=cost of the gold watch. 
$20=cost of the silver watch, and 
$140=cost of gold the watch. 

A man has two watches, and a chain worth $20; if he put 
the chain on the first watch it will be worth f- as much 
as the second watch, but if he put the chain on the sec- 
ond watch it will be worth 2| times the first watch 
what is the value of each watch? 

r 1. f s.=f f.+$20. 
2. i b.==J of (f f.+$20)=4 f.+$10. 
3.f s.=3 times (i f.+$10)=| f +$30. [lem. 

4. | s.= y f. — $20, by the second condition of the prob- 

5. .'•• V f — $20=| f.+$30, whence 

6. V f -— f f.==$30+$20, or 

7. | f.=$50. 

8. J f.=| of $50=$ 10, and 

9. 4 f_ 4 t j mes $10=$40=value of first watch. 
io. | s.=f f.+$30=| of $40+$30=$90=value of the sec- 
ond watch. 

$40=value of first watch, and 
$ ( ,)0=value of second watch. 

( White's Comp. Arith., p. 2J/.8, prob. 60 ) 

At the time of marriage a wile's age was | of the age of 
her husband, and 10 years after marriage her age was 
T ^ of the age of her husband ; how old was each at 
the time of marriage? 

1. |=husband's age at the time of marriage. Then 

2. |=wife's age at the time of marriage. 

3. H-10 years— husband's age 10 years after marriage. 

4. f+10 years=wife : s age 10 years after marriage. But 

5. jq-\-7 years= T 7 Tr of ( J-f-1 years)=wife , s age 10 years 
after marriage, by second condition of the problem. 

6. .'. i 7 6 I 7 years=$-|-10 years. Whence 

7. jq — 1=10 years — 7 years, or 

S. ,'(,=0 years. [of marriage. 

9. TJr=10 times 3 years=30 years=husband's age at time 

10. |, or T (; ff ,=6 times 3 years=18 years=wife's age at the 
time of marriage. 



II. 



II. , 



144 FINKEL'S SOLUTION BOOK. 

( 30 years=husband's age at time of marriage, and 
( 18 years=wife's age at time of marriage. 

( White's Conip. A., f. 2^1, firob. 35.) 

I. Ten years ago the age of A was J of the age of B, and 
ten years hence the age of A will be |- of the age of B ; 
find the age of each. 

1. |-=B's age 10 years ago. Then 

2. -j=A's age 10 years ago. 

3. J+10 years=B's age now, and 

4. f+10 years=A's age now. 

5. f+20 years=B's age 10 years hence, and 

6. f+20 years=A , s age 10 years hence. [hence. 

7. f of (f+20 years)=|+"l6| years=A's age 10 years 
II. < 8. .•. f+16f years=f+20 years ; whence 

9. f — f=20 years — 16f years, or 

10. TV=3i years, and 

11. T f=12 times 3^ years=40 years=B's age 10 years ago. 

12. f= T 9 2=9 times 3-J years=30 years=A's age 10 years 
ago. 

13. .". yf+10 years=50 years=B's age now, and 
-14. yV+10 years=40 years=A's age now. 

( 50 years=B's age, and 
^40 years=A's age. 

I. Two men start from two places 495 miles apart, and 
travel toward each other ; one travels 20 miles a day, 
and the other 25 miles a day ; in how many days will 
they meet? 

1. -|=nuinber of days. 

2. 20 mi.=distance first travels in 1 day. 

3. f X20 mi.=distance first travels in -| days. 

4. 25 mi.=distance second travels in 1 day. 

5. f X25 mi.=cHstance second travels in f days. 

6. .'. |X20 mi.+lx25 mi.=f x(20 mi.+25 mi.)=distance 
both travel. 

7. 495 mi.=distance both travel. 

8. .-. (20 mi.+25 mi.)xf=(45 mi.)Xf=495 mi. Whence 
^9. -|=495-J-45=ll=number of days. 

III. .-. They will meet in 11 days. 

Second solution. 

rl. 20 miles=distance first travels in a day. 
T1 I 2. 25 miles=distance second travels in a day. 

'I 3. .*• 45 miles=distance both travel in a day. [days. 

14. .-. 495 miles=distance both travel in 495-7-45, or 11, 



II. 



II. 



ANALYSIS. 145 

III. .-. They will meet in 11 days. 

Third solution — the one usually given in the schoolroom. 

20+25=45)495(11 days. 
45 
45 
45 

I. Find a number whose square root is 25 times its cube root. 

1. |— square root of the number. Then 

2. !xf=the number, because the square root X the square 
root equals the number. 

3. -|=the cube root of the number. Then 
4- '!X|Xf= the number. But 

5. -1=5 X(f)- Hence, squaring both sides, 

6. 4xf=25X(fXf). But 

7. |xf=the number, and 

8. f Xf Xf=the number. 
9- ■*. fXf X|=25X(|Xi). Dividing by (f Xf), 

10. f=25. 

11. ... (| )3=25 3 =15625. 

III. .-. The number is 15625. {R. H. A., p. 367,prob. 1!^.) 



I. A man bought a horse, saddle and bridle for $150 ; the 
cost of the saddle was \ of the cost of the horse, and the 
cost of the bridle was \ the cost of the saddle; what was 
the cost of each? 

1. i|=cost of the horse. Then 

2. -yy=\ of if=cost of the saddle, and 

3. Y2=\ of T 2 -2=cost of the bridle. 
4- lf=|i+ T \+ T V=costofall. 

IU5. $150=cost of all. 

6. .-. |4=$150, and 

7. T \= T K of $150=$10=cost of bridle. 
S. if=12 times $10=$120=cost of horse. 

19. T \=2 times $10=$20=cost of saddle. 



( $10=cost of the bridle, 
III. .'. < $20=cost of the saddle, and 
( $120=cost of the horse. 



( White's Comp. A., p. 2J f l,prob. 39.) 



146 



FINKEL'S SOLUTION BOOK. 



A and B perform y 9 -^ of a piece of work in 2 days, when, 
B leaving, A completes it in -J- day; in what time can 
each complete it alone? 



II.< 



III. .-. 



t 9 -q— part A and B do in 2 days. 
-fs—i of T 9 ^=part A and B do in 1 day. 
-1-2- — I 9 Q-= T 1 -Q=part left after B quits, and which A com- 
pletes in \ day. 
r 2 Q-=i=part A can do in 1 day. 
.-. -|=part A can do in -|-H==5 days. 
■£q — i=-^ r =J==part B can do in 1 day. 
.*. -|=part B can do in J-5-J, or 4, days. 

A can do the work in 5 days, and 
B can do the work in 4 days. 

( White's Comp. A., p. 280, prod. 193.) 



I. A and B can do a piece of work in 12 days, B and C in 9 
days, and A and C in 6 days; how long will it take 
each alone to do the work ? 



f 1. 
2. 
3. 



II. 



8. 
9. 

10. 
11. 

12. 

13. 

14. 



12 days=time it takes A and B to do the work. 
.-. T 1 Y=part they do in 1 day. 
9 days=time it takes B and C to do the work. 
.-. l=part they do in 1 day. 

6 days=time it takes A and C to do the work. 
.-. l=part they do in 1 day. 

... _i__|li._L.i. = ! |=part A and B, B and C, and A and C 
do in 1 day=twice the work A, B, and C do in 1 day. 



III. .• 



13 

72" 



f=-J of i|=part A, B, and C do in 1 day. 

T 7 2 = part A, B, and C do in 1 day — part B 



tV 



and C do in 1 day=part C does in 1 day. 
^-f=part C does in \\ I 7 7 2 , or lOf days. 
T f — i-= y 5 ¥ =part A, B, and C do in 1 day — part B and 

C do in 1 day=part A does in 1 day. 
■i|=part A does in ^|-^- T 5_=14|- days. ■ 
T § — i= T L=part A, B, and C do in 1 day — part A and 

C do in 1 day=part B does in 1 day. 
-!J-f=part B does in ^-|-f- T L=72 days 

i 14f days=time it takes A, 
< 72 days=time it takes B, and 
( 10f days=time it takes C. 

( White's Comp. A., p. 19^, prod. 280.) 



The head of a fish is 8 inches long, the tail is as long as 
the head and ■£ of the body-|-10 inches, and the body is 
as long as the head and tail ; what is the length of the 

fish? 



in 



ANALYSIS. 147 

'1. f=length of body. 

2. 8 in.=length of head. 

3. i 1. of b.+lO in.+8 in.=£ 1. of b.+18 in.=length of tail. 

4. | 1. of b.=length of head-|-length of tail. 

5. .-. f 1. of b.=(| 1. of b.+18 in.)+8 in.=J 1. of b.+26 in. 
Whence 

6. 2 1. of b.— -J 1: of b.==i 1. of b.=26 in. 
7- .'. f 1. of b., or length of body ,=2 times 26 in. =52 in. 
8. £ 1. of b.+18 in.=26 in.+18*V=44 in.=length of tail. 

.9. .*. 52 in. +44 in.+8 in.==104 in.=1ength of the fish. 

III. .-. The length of the fish is 104 inches. 

I. Henry Adams bought a number of pigs for $48 ; and 
losing 3 of them, he sold §- of the remainder, minus 2, 
for cost, receiving $32 less than all cost him; required 
the number purchased. 

1. f=remainder after losing 3. Then 
2 |-|-3=n umber at first. 

3. f of r. — 2=number sold. 

4. $48— $32=$ 1 6= \\ hat was received for § of r.— 2. 

5. $8=-£ of $16=\vhat was received for -£- of (f of r. — 2), 
or -J of r. — 1. 

(). $24=3 times $8=what was received for 3 times (- 1 - of 
r.— l)=f of r — 3. 

7. .-• $48— $24=$24=what (| of r.+3)— (| of r.— 3), or 6 
pigs cost. 

8. $4=4 of $24=what 1 pig cost. 

9. .-. $48=what 48-f-4, or 12, pigs cost. 

III. .-. lie bought 12 pigs. 

{Brooks' Int. A., f. 16& p>'ob. 9.) 

I. A bought some calves for $80; and having lost 10, he sold 
4 more than §■ of the remainder for cost and received 
$32 less than all cost; required the number purchased. 

1. |=rcmainder after losing 10. Then 

2. f-f-10=number purchased. 

3. f of r.-f-4=number sold. [cost. 

4. $80— $32=$48=cost of § of r.-f4, since they sold at 

5. $24=-J of $48=cost of J of (f of r.+4)=£ of r.+2. 
• 6. $72=:] times $24=cost of 3 times (£ of r.+2)=| of 

r.-f-6. [cost. 

7. .-. $80— $72=$8=what (| of r.+lO)— (| of r.+6), or 4 

8. $2=1- of $8=what 1 cost. 
19. $80=what 80-7-2, or 40 cost. 

III. .-. He bought 40 calves. 

(Brook's Bit. A., p. 16^, prod. 10.) 



n.\ 



148 'FINKEL'S SOLUTION BOOK. 

I. A lost |- of his sheep; now if he finds 5 and sells | of 
what he then has for cost price, he will receive $18; 
but if he loses 5 and sells | of the remainder for cost 
price, he will receive $6; how many sheep had he at 
first? 
( 1. |= the number of sheep he had at first. 

2. f= the number he lost. 

3. | — f=f» tne number he had after losing- -|. 

4. f+5= the number he had after finding 5. 

5. f of (|+5)=2 6 5+3, the number he sold. 

6. 4 — 5= the number, had he lost 5. 

7. f-of(f — 5)=2 6 5 — 3, the number he would have sold. 

8. $18=what (2 6 5+3) sheep cost. 
II.<J 9. $6= what (^— 3) sheep cost. 

10. .'. $12=$18— $6=what (2VKO sheep-— (-fe—S) sheep, 
or 6 sheep cost. 

11. $2=1 of $12= what 1 sheep cost. 

12. $18= what 18-T-2, or 9 sheep cost. But 

13. $18=what (2 6 s+3) sheep cost. 

14. •"•-^5+3 sheep = 9 sheep, or 

15. -2 6 5-=6 sheep. 

16. yV— 6 °^^ sheep=l sheep, and 

17. f|=25 times 1 sheep =25 sheep. 
III. .'. He had 25 sheep at first. 

(BrooPs hit. A., p. 165,prob. 15.) 

I. A man bought a certain number of cows for $200; had he 
bought 2 more at $2 less each, they won id have cost 
him $216; how many did he buy ? 

1. $200=cost of cows. 

2. $216=cost of oiiginal number of cows-j-2 more. 

3. $216— $200=$16=cost of 2 cows at $2 less per head. 
■ 4. .-. $8=i of $16=cost of 1 cow at $2 less per head. Then 

5. $8+$2=$10=cost of each cow purchased. 

6. $200=cost of 200—10, or 20 cows. 
III. .*. He bought 20 cows. 

(Brook's Int. A., p. 162, prob. 8.) 

I. A person being asked the hour of day, said, "the time 
past noon is -^ of the time past midnight;" what was 
the hour? 



IU 



II. 



1. |=time past midnight. 

2. 



3. .'. -| — i=|=time from midnight to noon. 



4. 12 hours=time from midnight to noon. 

5. .". f=12 hours. 

6. j=i of 12 hours=G hours=time past noon. 
III. It was 6 o'clock, P. M. 



ANALYSIS. 



149 



iM 



in. 
i. 



ii. 



Provided the time past 10 o'clock, A. M., equals f of the 
time to midnight; what o'clock is it? 

1. |-=time to midnight. Then 

2. f=time past 10 o'clock. 
±- |-f =|=time from 10 o'clock to midnight. 
14 honrs=time from 10 o'clock to midnight. 
.'. |=14 hours. 

i=Y of 14 hours=2 hours, and [o'clock P. M. 

f— 3 times 2 hours=6 hours, time past 10 o'clock=4 

.-. It is 4 o'clock, P. M. 

At what time between 3 and 4 o'clock will the hour and 

minute hands of a watch be together? 

!==distance the h. h. moves past 3. Then 



2 ^ 4 =12X^=di s tance the m. h. moves past 12. 
2 -£ — |-= 2 -2 2 :=distance the m. h. gains on the h. h. 



15 min.=distance the m. h. gains on the h. h 



.-. 2 2 2 =15 rain. 

$=-£% of 15 tnin.=44 rain. 



2_4 = 24 times 



2 2 



[past 12. 



:16y T min.=distance m. h. moves 



III. .-. It is 16 T 4 T min. past 3 o'clock. 

Remark. — In problems of this kind, locate the minute hand at 12 and the 
hour hand at the first of the two numbers between which the conditions of 
the problem are to be satisfied. Thus in the above problem, at 3 o'clock 
the minute hand is at 12 and the hour hand at 3. 

The minute hand moves over 60 minute spaces while the hour hand 
moves over 5 minute spaces. Hence the minute hand moves 12 times 
as fast as the hour hand. Since at 3 o'clock the minute hand is at 12 
and the hour hand at 3, and the minute hand moves 12 times as fast as the 
hour hand, it is evident that the minute hand will overtake the hour hand 
between 3 and 4. So we let |=distance the hour hand moves past 3 until 
it is overtaken by the minute hand. But since the minute hand moves 12 
times as fast as* the hour hand, while the hour move §, the minute hand 
moves 12 times |, or 2 2 4 - . Now the minute hand has moved from 12 to 3+|, 
or 15 minutes-4-|. Hence the minute hand has gained 15 minutes on the 
hour hand. It has also gained 2 2 4 — \, or ? f 2 - ■'• z f=l5 minutes. 

In solving any problem of this nature, first locate the hands as previously 
stated, and then ask yourself how far the minute hand must move to meet 
the conditions of the problem, if the hour hand should remain stationary. 

I. At what time betweeu 6 and 7 o'clock will the minute 
hand be at right angles with the hour hand? 

1. 2 f =distance h. h. moves past 6. 

2. 2 ^=12 times |=distance m. h. moves past 12. 
3 _•_ 2j4 — 2__2_2__ c |j s ^ ance 1T1 ]-, gains on h. h. 

4. 15 min. or 45 min.=distance m. h. gains on the h. h. 

5. .'. 2 ^=15 min. or 45 min. 

6. $=ys of 15 min. or ■£% of 45 min.=H min. or 2. r 1 o min. 

7. 2 ^=24 times ^-| min. or 24 times 2-^V min.=lG T j min. 
or 49y T min. 

.-. The minute hand will be nt right angles with the hour 
hand at 16 T 4 T min. or 49^ min. past 6 o'clock. 



II. 



III. 



150 



FINKEL'S SOLUTION BOOK. 



Explanation. — Locate the minute hand at 12 and the hour hand at 6. Now 
if the hour hand had remained stationary at 6, the minute hand would have 
to move to 3 or 9, i. e., it would have to gain 15 min. or 45 min. While the 
minute hand is moving to 3 the hour hand is moving from 6. So the min- 
ute hand must move as far past 3 as the hour hand moves past 6. Or while 
the minute hand is moving to 9 the hour hand is moving past 6. So the 
minute hand must move as far past 9 as the hour hand is past G. .". The 
minute hand must gain 15 minutes in the first case and 45 minutes in the 
second. 

I. At what time between 2 and 3 o'clock are the hour and 
minute hands opposite? 



II. 



III. 



1. 

2. 

3. 

4. 

5. 

6. 
17. 



•|=distance hour hand moves past 2. Then 
2 ^=distance the minute hand moves past 12, in the 

same time. [hand. 

.'. 2 ^ — |= 2 _2__ -distance minute hand gained on the hour 
40 min.=distance the minute hand gained on the hour 

hand. 
.-. 22=40 min. 



i i 

2 2"2~ 

24 
2 



of 40 mim.= 
24 times 1 T 9 T 



It is 
opposite 



43 T \ 



=1- 



pa 



r min., and 
=43 T 7 T min. 

t 2 o'clock 



when the hands are 



Explanation. — Locate the minute hand at 12 and the hour hand at 2. Now 
if the hour hand remained stationary at 2, the minute hand would have to 
move to 8 or over 40 minutes in order to be opposite the hour hand But 
while the minute hand is moving to 8, the hour hand is moving from 2. So 
the minute hand must move as far past 8 as the hour hand is past 2. Since 
§ is the distance the hour hand moves past 2, f must be the distance the 
minute hand must move past 8. Hence the distance the minute hand 
moves is f+40 min. But 2 2 4 =distance the minute hand moves. .'. V = H~ 
40 min. or ^==40 min. as shown in step 5. 

I. At w T hat time between 3 and 4 o'clock will the minute 
hand be 5 minutes ahead of the hour hand? 



II. 



distance hour hand moves while the 
to be 5 min. ahead. 



h. is moving 
[moves J. 
h. 



2 ^=12 Xf=distance minute hand moves while the h 

.'.'?£ — |= 2 Y 2 =distance gained by the minute hand. 

15 min.-f-5 mim.=20 min.=distance gained by the m 

.-. % 2 =20 min. 

-J=T2 °f 20 min.=if min. 

2 2 4 =24 times if min.=21 T 9 T min. 



h. 



III. .'.It is 21^ min. past 3 o'clock. 



Explanation. — Locate the minute hand at 12 and the hour hand at 3. 
Now if the hour hand remained stationary at 3, the minute hand would 
have to move to 4 in order to be 5 min. ahead. But while the minute hand 
is moving to 4 the hour hand is moving from 3. Hence the minute hand 
must move as far past 4 as the hour hand moves past 3. But the hour hand 



ANALYSIS. 



151 



moves § past 3; hence, the minute hand must move -f+5 min. past 4, in all, 
|+20 min. Hence, the minute hand gains (f+20 min.) — §=20 min. on the 
hour hand. 

Remark. — We always find \ 4 , the distance the minute hand moves, for it 
indicates the time between any two consecutive hours. The hour hand 
indicates the hour. 



I. 



II. 



III. 



At what time between 4 and 5 o'clock do the hands of a 
clock make with each other an angle of 45° ? 

1. |=distance the hour hand moves past 4. 

2. 2 #=distance the minute hand moves past 12. 

3. .'. 2 2 4 — f= 2 "2 2 = distance the minute hand gains on the 

hour hand. 
12-^ min. or 27-J min.=distance gained by minute hand. 

[min. 
i. or U 



:12-|- min. or 27-g- min. 



%=Y2 ot 12J min. or -j^ of 27-J min.: 



44 



=24 times \\ min. or 24 times \\ min.=13 T 7 T 
or 30 min. 



At 13^ min. past 4 or 30 min. past 4, the hands make 
an angle of 45° with each other. 

Explanation. — Locate the minute hand at 12 and the hour hand at 4. 45° 
=| of 360 n . I of GO min.=7£ min. Hence, that the hands make an angle of 
45°, the minute hand must be either 7£ minutes behind the hour hand or ~\ 
min. ahead, Now if the hour hand remained stationary at 4, the minute 
hand would have to move over 12£ min. or 2£ min. past 2. But 
while the minute hand is moving this distance, the hour hard is moving 
past 4. Hence, the minute hand must move as far past 2| min. past 2 as the 
hour hand moves past 4, i. e., the minute hand moves |-j-12£ min. Hence, 
it gains (|4-12£ min.) — §=12^ min. The reasoning for the second result is 
the same as for the first. 



II. 



At what time between 4 and 5 o'clock is the minute hand 
as far from 8 as the hour hand is from 3? 

1. §=distance the hour hand moves past 4. 

2. 2 ^=12 times |=distance minute hand moves past 
12 in the same time. 

3. .'. V4~r> — ~> (; — Stance both move. 

4. 35 min.=distance both move. 



LB. 



III. 



6. i 



:35 min. 



2 

:Jg- of 35 min.=l^ min. 
—24 times 1^- min.^32-^ min. 



1. f=distance the h. h. moves past 4. 

2. 2 ^ 4 =distance minute hand moves past 12. 

3. .*. 2 j — r r ='Y=distance the minute hand gains. 

4. 45 min.=distance the minute hand gains. 

5. .-. 2 Y 2 =45 min. 

6. -£=^>- of 45 min. =2o 1 T T min. 

7. 2 2 4 ==24 times 2j\ min"— 49^ min. 



It 



is32A 



min. or 49 T y min. past 4 o'clock. 

(R> H. A.,p. b.08,$rob. 4.0.) 



152 



FINKEL'S SOLUTION BOOK. 



Explanation. — This problem requires two different solutions. Locate the 
minute hand at 12 and the hour hand at 4. The hour hand is now 5 min- 
utes from 3. If the hour hand remained stationary, the minute hand would 
have to move to 7 to be 5 minutes from 8. But while the minute hand is 
moving to 7, the hour hand is moving past 4. Hence the minute hand must 
stop as far from 7 as the hour hand moves past 4; i. e., if the hour hand 
moves § past 4 the minute hand must stop \ from 7. Then the hour hand 
will be 5 minutes-|-f from 3 and the minute hand will be f-)-5 minutes from 
8. While the hour hand moved -§, the minute hand moved 35 min — § .*. *g 
=35 min. — §, whence \ 6 =35 min. .-. 35 min.=distance they both move. 
The second part has been explained in previous problems. 



At what time between 5 and 6 o'clock is the minute hand 
midway between 12 and the hour hand? When is the 
hour hand midway between 4 and the minute hand? 



fl. f=distance the hour hand moves past 5. 
2. 2 #=distance the minute hand moves in 



the 



same 



time. 



-|— f-25 min.=distance from 12 to the hour hand. 



II. 



3. 

4. -J- of (f-|-25 min.)=^~)-12-| min.=distance minute 
hand moves. 

:-H-12£ min. 

min. 



2_4. 
2 " 



.25 
■^6" 



min. 



IB. 



i=^ of l 2 i min - : 
2 2 4 =24 times ff min.=13^ min - 
|^=:distance the hour hand moves past 5. 
2 ^=distance the minute hand moves in the 
time 



same 



f-|-5 min.=distance the hour hand is from 4. 

^— j— 10 min.=2 times (f-f-5 min.)=distance the min- 
ute hand is from 4, since the hour hand is midway 
between it and 4. 

20 min.-f(f-f-10 min.)=|-f30 min.=distance the 
the minute hand is from 12. 

... 2_4 = 4_J_30 min ? or 

2 ^— f= 2 2°=30 min. 
i=-^Q of 30 min.=l-2- min. 
2_4 = 24 times 1^ min.=36 min. 

It is 13^=3 min. past 5 o'clock. 
It is 36 min. past 5 o'clock. 

(R. H. A., f. 40S,pro6. J/.1.) 

Explanation. — Locate the minute hand at 12 and the hour hand at 5. If 
the hour hand remained stationary, the minute hand would have to move 
over \ of 25 minutes, or 12^ minutes. But while it is moving over 12| 
minutes, the hour hand is moving past 4. Hence, the minute hand will 
have to move 12£ minutes+^ of the distance the hour hand moves past 4. 
Hence - 2 /:=£4-12£ minutes, as shown by step 5 of A. In B, if the hour hand 
remained stationary, the minute hand would have to move over 30 minutes, 
i. e., to 6, that the hour hand may be midway between it and 4. But while 
the minute hand is moving to 6 the hour hand is moving past 4. Hence 
the minute hand must move twice as far past 6 as the hour hand moves past 



III. 



6. 

7. 
8 
9. 

A 

B. 



ANALYSIS. 



153 



§=distancethe hour hand moves past 4; hence, -f= 
Hence, 1+30 minutes=distnnce 



4. But 

ufe hand moves past 6 



distance the min- 
the minute hand 



. -2-4 = |-f 30 minutes, as shown by step 6 of B. 



At what time between 3 and 4 o'clock will the minute 
hand be as far from 12 on the left side of the dial plate 
as the hour hand is from 12 on the right side? 



II. 



:distance the hour hand moves past 3. 



2 2 4 =12 times 



i-=distance the minute hand moves in "the 



same time. 
2 ^ 4 -|-£= 2 2 6 =distance they both move. 
45 min.=di stance they both move. 
.-. 2 or 6 =45 min. 

min. 



4=^ of 45 mm. 



.1 1 9 



2 4 = 24 times 1M mm 



41 



l 3 



min. 



III. .-. It is 41 T ^ min. past 3. 



Explanation. — Locate the minute hand at 12 and the hour hand at 3. If 
the hour hand remained stationary, the minute hand would have to move to 
9 to be as far from 12 on the left side of the dial plate as the hour hand is 
from 12 on the righi. But while the minute hand is moving to 9, the hour 
hand is moving past 3. Hence, the minute hand must stop as far from 9 as 
the hour hand moves past 3. Hence, it is evident, they both move 45 
minutes. 



I. 



II. 



A man looked at his watch and found the time to be be- 
tween 5 and G o'clock Within an hour he looked 
again, and found the hands had changed places. What 
was the exact time when he first looked? 

(1.) |=distance m. h. was ahead of h. h., or the dis- 
tance the h. h. moved, since it changed place 
with the m. h. [the two observations. 

2 ^=distance the m. h. moved in the time between 
/. 2 ^-|-§-= 2 2 6 =distance they both moved. 
60 min.=distance they both moved. 
.-. 2 2 6 =60 min. 

%=Yt °f ^0 min.=2 T \ min. [ahead of h. h. 

f=2 times 2 T 4 -j min.=4 T 8 ^ min.=distance m. h. was 
|=distance h. h. was past 5, at time of first obser- 
vation. Then [servation. 
2 ^=distance m. h. was past 12 at time of first ob- 
25 min.-f-f-f-4 T \ min.=f+29 1 ^= distance m. h. 
was past 12 at time of .first observation. 

¥-f=V=29A min. 
+=,-'-, <>f29 T % min.=l^mm. 



(2. 
(3. 
(4. 

A 5 

(6. 



(7-) 



(S. 



:24 times 1-^- min.=32 T 4 3 min. 



III. /. It was 32/;.- min. past 5 o'clock. 



154 



FINKEL'S SOLUTION BOOK. 



Explanation. — It is clear that the minute hand -was ahead of the hour 
hand at the time of the iirst observation, or else they could not have ex- 
changed places within an hour. Now, we call the distance from the point 
where the hour hand was located at first to the point where the minrte hand 
was located first, f. But in the mean time the hour hand has moved to the 
position occupied by the minute hand and the minute hand has moved on 
around the dial to "the position occupied by the hour hand, i. c, the hour 
hand has moved \ and the minute 12 times f, or \ 4 . Hence, they both 
moved 2 2 6 . They both moved 60 minutes since the hand moved on around 
the dial to the position occupied by the hour hand and the hour hand mov- 
ed to the position occupied by the minute hand. .' 2 2 6 =60 min. as shown 
in step (5.) The remaining part of the solution has been explained in pre- 
vious problems. 

I. At a certain time between 8 and 9 o'clock a bov stepped 
into the schoolroom, and noticed the minute hand be- 
tween 9 and 10. He left, and on returning within an 
hour, he found the hour hand and minute hand had ex- 
changed places. What time was it when he first en- 
tered, and how long was he gone? 



rA. 



II. 



III. 



B. 



((1.) f=di stance m. h. was ahead of the h. 
tance it moved. 
(2.) 2 2 4 = distance m. h. moved while the h. 
(3.) 2 2^-|-f= 2 Y=di stance both moved. 



h. or dis- 

[f- 

h. moved 



(4) 

(5.) . 

(6.) i 

(7-) I 



1(8.) 



2 

60 min =distance both moved. 
.-. 2_6 ==( 3o min. 

[was ahead 
min.=distance m h. 
distance h. h. moved past 8 
=distance m. h. moved in same time. 



of 60 min.=2 T \ min. 



:2 times 2 T \ min. 



:4y 8 



2_ 

2"— 

24. 
2" 

40 



-1+44^ 



h. moved to be 4 T 8 ¥ 



min -hf+4 T 8 -g- min. 
tance m 
.-. 2 2 4 =H-44 T % min 

2 4 2 2 2 A A 8 

2 ^ — 2" — **T1 
i=2 T 2 ° f 44 A m 
2 T 4 =24 times 

2 2- 6 =distance they both moved. 
60 min.=distance they both moved. 
.-. 2-^=60 min. 



min. = dis- 



ahead. 



A m]n - 




mir..=2 T f^ min. 


[past 8. 


2 T ^ min. = 48^ 


min.=time 



^ of 60 min.= 



, 4 =24 



times 2 T 4 3 



-9 4 

-z T 



mm. 



[was gone. 
min.=time he 



A. It was 48 T 9 ¥ 6 g- min past 

tered school room. 

B. He was gone 55 T \ min. 



o'clock when he first en- 



Suppose the hour, minute, and second hands of a clock 
turn upon the same center, and are together at 12 
o'clock ; how long before the second hand, hour hand, 
and minute hand respectively, will be midway between 
the other two hands? 



ANALYSIS. 



L55 



II. 



B. 



Ic. 



5. 



8. 

9. 
10. 
11. 

1. 
2 
3. 

4. 



6. 

7. 

8. 

9. 
10. 
11. 

1 
2. 



|-— distance the hour hand moves past 12. Then 
2 ^=di stance the minute hand moves past 12, and 
i-4_4o == 720 times f=distance T 

the second hand moves past 

12. 

= distance 



1 44 2 4 1 4_1 

~2~ 2 2 

from the minute hand to the 



second hand. 



distance 



1440 2_4 1_4J_G 

2 2 2 

from the second hand to the 




TIG.L 



hour hand. 

2 ^ — \ = 2 ^ 2 = distance from 
the hour hand to the second hand. 
i_4_i_6^i^L6_j_2^ == 2_^5_4 == distance around the dial. 
60 seconds=distance around the dial as indicated 
by one revolution of the s. h. 
... 2jyL4 ====6 o sec. 

sec. 



WoT of 60 sec - 



3 
T4~2T 



L4_4.j = i440 times T ff T sec. =30^7- sec. = time 
when s. h. is midway between the h. h. and m. h. 
|=distance the hour hand moves past 12. Then 
2 2*=distance the minute hand moves past 12, and 



M^-0 = distance tne 

hand moves past 12 

24 
2 

h. to m. h. 



distance the second 



|= 2 ^==di stance from h. 




Y=distance from s. h. to h. 
h., because the h. h. is mid- 
way between them. [12. 

2 ^ 2 — |— distance from s. h. to 

i-4p-py> = 1JHLP = distance 

arouiuf the dial. rIG.£>. 

60 sec.=distance around the dial. 

.-. U_6_o = (30 sec. 

£=T?Tf o of 60 sec.= f \ sec. 

L^li.^1440 times ^ sec.=59|| seedtime when 
the h. h. is midway between the s. h. and m. h. 

-|=distance h. h. moves past 12. Then 
h. moves past 12, and 



2 /=distance m. 



lJLLP 
2 



distance s h. moves 
past 12. [h. to s. h. 

! 2 4 — != 2 -2 2 =distance from h. 
^ 2 =distance from m h. to 
s. h. [from 12 to s h. 

_L2_2 _|_ 2^2 = 4_e = distance 
/ = 1 -^^ = distance 
around the dial. [dial. 

60 sec =distance around the 



2 
•> 

14- 




FIG 3. 



156 



FlNKEL'S SOLUTION BOOK. 



9. 
10. 
11. 



A. 



III. .-. 



1 B c. 



.-. i3_9 4 =z go sec. 

i=TWWZ ° f 60 SeC -=(fW SeC - 

14 Y 40 =1440 times -g 3 g° T sec.=61|~§-§- sec.=time past 
12 when the m. h. will be midway between the 
h. h. and s. h. 

The second hand is midway between h. h. and m. 

h. at 30 T 3 ¥ 9 2° T sec. past 12. [at 59f| sec. past 12. 
The hour hand is midway between s. h. and m. It. 
The minute hand is midway between h. h. and s. 

h. at 61ff? sec. past 12. 



Explanation. — A. We represent the distance moved by the hour hand 
by |, = the space Th. And since the minute hand moves 12 times as fast 
as the hour hand, it moves %£. The second hand moves 60 times as fast as 
the minute hand or 720 times as fast as the hour hand. From T to h is \ 
and from / to m is %g. .'. From h to m is Tin — Th - 2 / — | = \*. From T to 
s is 14 2 40 - •'• From m to s = Ts—Tm = iyo__y - lye. And, by the condi- 
tion of the problem, the distance from m to .s = the distance from m to h. .'. 
from m to h = 1 V 6 + 1 V 6 = 2 Y 2 - We have seen, already, that the distance 
from // to m is 2 ^. .'. The whole distance around the dial is 28 2 32 + 2 2 2 = 28 2 54 . 

B. From T to h is f. From T to m is V. /. From h to m=Tm—Th= 
2 2 4 — f =2 5 ? . By the condition of the problem, the distance from h to 7ra=the 
distance from .9 to //. :.sT — Tk=£$- — f= 2 2°. From T around the dial to 
the right of s is 1 Y°- •"• The whole distance around the dial= 1 Y°+- 2 2 = 

14_6 ' . 

C. From T to h is |. From 7 1 to w is V- •'• From £ to m=*£—\—Z£. 

By the condition of the problem, the distance from m to snathe distance 
from h to m=*g. .'. From T to s is f+^+fg 2 '— V- From T around the 
dial through T to j is tA ^ & . .'. The whole distance around the dial is J-y- - — 



lU 



III. 



A sold to B 9 horses and 7 cows for $300 ; to C, at the 
same price, 6 horses and 13 cows, for the same sum; 
what was the price of each? 

1. Cost of 9 horses+cost of 7 cows=$300. Then the 

2. Cost of 36 horses+cost of 28 cows=$1200, by taking 

4 times the number of each. 

3. Cost of 6 horses+cost of 13 cows=$300. Then the 

4. Cost of 36 horses -(-cost of 78 cows=$1800, by taking 

6 times the number of each. But 

5. Cost of 36 horses+cost of 28 cows=$1200. 

6. .*. Cost of 50 cows=$600, by subtracting; and 

7. Cost of 1 cow= 1 1 j of $600=$12. The 

8. Cost of 7 cows=7 times $12=$84. 

9. .-. Cost of 9 horses=$300— cost of 7 cows=$300- 

=$216. The 
10. Cost of 1 horse=i of $216=$24. 

( The cows cost $12 apiece, and 
' " ( The horses $24 apiece. 



ANALYSIS. 



157 



I. 



II. 



A man at his marriage agreed that if at his death he 
should leave only a daughter, his wife should have ^ of 
his estate; and if he should leave only a son she should 
have •£. He left a son and a daughter. What fractional 
part of the estate should each receive, and what was 
each one's portion, if his estate was worth $6591? 
|-=daughter's share. 



|=wife's share. 

9. 



[estate. 
of whole 



. l 
rs; 

T 3 -g- of whole es 



3 times f=son's share. 
J_|_j_|_! == i T s = _ w hole estate. 
$6591=whole estate. 
/. y>=$6591. 

7. _i =T i_ of $G591=$507=daughter , s share, 

8. |=3 times $507=$1521=wife's share, 
tate. [tate. 

^9. 9=9 times $507=$4563=son , s share,= T 9 -j of whole es- 
( $507=y 1 ^ of whole estate— daughter's share. 
III. .-. < $1521= T \ of whole estate=wife's share. 
( $4563= T % of whole estate=son's share. 

(Millies Prac. A., p. 362, prod. 7£.) 

Note. — For a valuable critique, by Marcus Baker, U. S. Coaj>t Survey, on 
this class of problems, see School Visitor, Vol. IX., f. 1S6. 

I. There is coal now on the dock, and coal is running on 
also from a shoot at a uniform rale. Six men can clear 
the dock in 1 hour, but 11 men can clear it in 20 min- 
utes ; how long would it take 4 men? 

1. |=what one man removes in 1 hour. Then 

2. 1 2 2== 6 times |=what 6 men remove in 1 hour. 

3. f= 1 }- of |=what 1 man removes in 20 min., or J hour. 



II. 



6. 

7. 
8. 
9. 

10. 



:11 times f=what 11 men remove in ^ hour 



Th 



— 2^=y= what 



runs on in 



en 



J= 1 -g*-s-f=what runs on in 1 hour. 



i 9. 



1 hr. — J hr.=| hr 



[commenced, 
when the work 



HI. 



|=!=what was on the dock 
|=what 4 men remove in 1 hour. 
.•. § — £=*|==part of coal removed every hour, that was 

on the dock at first. 
|=coal to be removed in |-f-J=5 hours. 

(R. H. A., p. £06, prod. 90.) 
It will take 4 men, 5 hours to clear the dock. 



Explanation. — y — what 6 men remove in 1 hr. and \? 



what 1 1 men re- 

1 2 2 2 — 14 

2 6 ft 

amount of coal that ran on the dock from the shoot in 1 hr. — £ hr , or jj- hr. 
Hence in I hr. there will run on, V~*~§~ I- Since \ run on in 1 hr. and y 
:=the whole amount of coal removed in 1 hr., y — |, or § must be the 
amount of coal on the dock when the work began. Since |— the amount 4 
men remove in J hr. and X=the amount that runs on the dock in ] hr., § — 
\, or I is the part of the original quantity removeo each hour. Hence, if £ 



is removed in 1 hour | would be removed in £-r- 



hours. 



158 FINKEL'S SOLUTION BOOK. 

I. If 12 oxen eat up 3-J- acres of pasture in 4 weeks, and 21 
oxen eat up 10 acres of like pasture in 9 weeks ; to find 
how many oxen will eat up 24 acres in 18 weeks. 

1. 10 parts (say)=what one ox eats in a week. Then 

2. 120 parts=12xl0 parts=what 12 oxen eat in 1 week, 

3. 480 parts=4xl20 parts=what 12 oxen eat in 4 weeks. 

4. .'. 480 parts=original grass-f-growth of grass on 3-J- A. 
in 4 weeks. 

5. 144 parts—— of 480 parts=original grass-|-growth of 
of 

grass on 1 A. in 4 weeks. 

6. 210 parts=21XlO parts=what 21 oxen eat in 1 week, 

7. 1890 parts=9 X210 parts=what 21 oxen eat in 9 weeks. 

8. .'. 1890 parts=original grass-[-growth of grass on 10 
A. in 9 weeks. 

9. 189 parts= T L of 1890 parts=briginal grass-j-growth 
on 1 A in 9 weeks 

H-< 10. .'. 189 parts — 144 parts=45 parts=growth on 1 A. in 
9 weeks — 4 weeks, or 5 weeks. 

11. 9 parts=^ of 45 parts=growth on 1 A. in 1 week. 

12. 36 parts=4x9 parts==growth on 1 A. in 4 weeks. 

13. .*. 144 parts — 36 parts=108 parts=original quantity of 
grass on 1 A. 

14 2592 parts=24xl08 parts=original quantity on 24 A. 
15. 216 parts=24x9 parts=growth on 24 A. in 1 week. 
16 3888 parts=18x216 parts=growth on 24 A. in 18 
weeks. 

17. .-• 2592 parts+3888 parts=6480 parts=quantity of 
grass to be eaten by the required oxen. 

18. 180 parts=18X10 parts=what 1 ox eats in 18 weeks. 

19. .• 6480 parts=what 6480-^-180, or 36 oxen eat in 18 
weeks. 



III. .-. It will require 36 oxen to eat the grass on 24 A. in 18 
weeks. 



Note. — This celebrated problem was, very probably, proposed by Sir 
Isaac Newton and published in his Arithmetica Universalis in 1704. Dr. 
Artemas Martin says, "I have not been able to trace it to any earlier work." 
For a full treatment of this problem see Mathematical Magazine, Vol. 1, 
No. 2. 



A man and a boy can mow a certain field in 8 hours, if 
the boy rests 3f hours, it takes them 9-J- hours. In what 
time can each do it? 



II. 



ANALYSIS. 159 

1. 9£ hr.~— 3| hr.=5f hr.=time they both work together in 

the second case. 

2. 8 hr.=time it takes them to do the work. 

3. .\ -J=part they do in 1 hour. 



51 2 3. 



:5f times ^=part they do in 5f hours. 



5. •'. f | — § |=^ 9 2 =part the man did in 3f hours, while the 
boy rested. 

6. .:. ^=— of -3 9 2=part the man did in 1 hour. 

7. .-. 4o=_p ar t the man can do in %% ; 4 3 or 13-J- hours. 

8. -£ — 1 p (J =g 1 Tr =part the boy does in one hour. 

9. '. |J=part the boy can do in gj-s-^. or 20 hours. 
TTT I It will take the man 13-^ hours, and 

•"• I The boy 20 hours. (R. H. A., p. 1+02, frob. 30.) 

I. Six men can do a work in 4-J days; after working 2 days, 
how many must join them so as to complete it in 3f 
days ? 

1. 4^ days=time it takes 6 men. 

2. 26 days=6 times 4^ days=time it takes 1 man. 

3. .". ^g-=part 1 man does in 1 day. 

4. j § j=6 times 2 1 TI =part 6 men do in 1 day. 

5. xV=2 times 1 :i . f =part 6 men do in 2 days. [days. 
<J 6. yf — T 6 ^= T 7 £=part to be done in 3| days — 2 days, or 1-| 

li 

7. ~==jljf==:pa.rt 1 man does in 1| days, 
zb 

8. .*. ^=part t^-ttJt' or ^ men can c ^° * n ^1 days. 

9. .'. 10 men — 6 men===4 men, the number that must join 
them. 

III. /. They must be joined by 4 more men that they may com- 
plete the work in 3| days. J?. H. A., p. ^02, prod. &£. 

I. From a ten-gallon keg of wine, one gallon is drawn off 
and the keg filled with water; if this is repeated 4 
times, what will be the quantity of wine in the keg? 
'1. T 1 (y =part drawn out each time. 

2. T 9 =part that was pure wine after the first draught. 

3. j - of tV ~ i 8 P ar<: w i ne drawn off the second draught. 

4. T 9 y — T f- g = 1 Yo P ai 't pure wine left after the second 
draught. [draught. 

5. T V of T 8 J r == 1 |J- - = part wine drawn off at the third 

6. yVy — T"!ro = iWu = P art P ure wine left after the third 
d r a ugh t. [ d r a u g h t . 

7. tV °^" l'lnru— rilu i)~ ! )art wine drawn off at the fourth 
*• Jifa\~UU^isffifa== P art P ure wine left after fourth 

draught. [fourth draught. 

9. .'. xVoVo °^ 1® gal. =6. 561 gai".=pure wine left after the 



II. 



160 FINKEL'S SOLUTION BOOK. 

III. .'. There will be 6.561 gal. of pure wine in the keg after 
the fourth draught. 

I. In the above problem, how many draughts are necessary 
to draw off half the wine? 

1. y L=part wine drawn off at the first draught. 

2. y£ — T L= T 9 ¥ =part wine left after the first draught. 

9 

3. ^q of t 9 -q= t | ¥ =— — =part wine drawn off at the sec- 
ond draught. 

92 ~ 92 

4. T \ — — -=-^ u =( — ) =part wine left after the second 
10 10 

draught. 

9292 

5. y 1 ^ of (— ) =— — =part wine drawn off at the third 

[I <J draught. 

9 2 92 93 

6. (— ) — -^— -=(— ) =P art wine left after tlie third 

draught. By induction, 

7. ( T 9 o-) n =part wine left after the nth draught. 

8. .'. 10( T 9 Q-) n =number of gal. left after the nth draught. 

9. 5— number of gal. left after the nth draught. 

10. .*. 10( T 9 o) n =5, whence 

11. ( T 9 Q) n =i Applying logarithms, 

12. n log. T V=log. \. 

13. .-. n = log. |^-log. T %=.30103-^. T".954243=.301030-r- 
.045757=6+. 

III. .". In 7 draughts, half and a little more than half of the 
wine will be drawn off. 



PROBLEMS. 

1. A man bought a horse and a cow for $100, and the cow 
cost §- as much as the horse; what was the cost of each? 

Ans. horse, $60; cow, $40. 

2. Stephen has 10 cents more than Marthia, and they to- 
gether have 40 cents; how many have each? 

Ans. Stephen, 25/; Marthia, 15/. 

3. A's fortune added to \ of B's fortune, equals $2000; what 
is the fortune of each, provided A's fortune is to B's as 3 to 4? 

Ans. A's, $1200 ; B's, $1600. 

4- If 10 oxen eat 4 acres of grass in 6 days, in how many 
days will 30 oxen eat 8 acres? Ans. 4 days. 



PROBLEMS. 161 

5. If a 5-cent loaf weighs 7 oz. when flour is worth $6 a bar- 
rel, how much ought it weigh when flour is worth $7 per barrel ? 

Ans. 

6. A lady gave 80 cents to some poor children; to each boy 
she gave 2 cents, and to each girl 4 cents; how many were there 
of each, provided there were three times as many boys as girls? 

Ans. 8 girls; 24 boys. 

7. Two men or three boys can plow an acre in \ of a day ; 
how long will it take 3 men and 2 boys to plow it? 

Ans. yL da. 

8. A agreed to labor a certain time for $60, on the condition 
that for each day he was idle he should forfeit $2, at the expira- 
tion of the time he received $30; how many days did he labor, 
supposing he received $2 per day for his labor? Ans. 22-J- days. 

9. The head of a fish is 4 inches long, the tail is as long as 
the head, plus \ of the body, and the body is as long as the head 
and tail ; what is the length of the fish? A?zs. 32 inches. 

10. In a school of 80 pupils there are 30 girls; how many boys 
must leave that there may be 3 boys to 5 girls? Ans. 38. 

11. A steamboat, whose rate of sailing in still water is 12 
miles an hour, descends a river whose current is 4 miles an hour 
and is gone 6 hours; how far did- it go? Ans. 32 miles. 

12. A man keeps 72 cows on his farm, and for every 4 cows 
he plows 1 acre, and keeps 1 acre of pasture for every 6 cows; 
how many acres in his farm? Ans. 30 acres. 

13. A company of 15 persons engaged a dinner at a hotel, but 
before paying the bill 5 of the company withdrew by which each 
person's bill was augmented $^; what was the bill? Ans. $15. 

14. A man sold his horse and sleigh for $200, and |- of this is 
8 times what his sleigh cost, and the horse cost 10 times as much 
as the sleigh ; required the cost of each. 

A?is. horse, $200; sleigh, $20. 

15. A went to a store and borrowed as much as he had, and 
spent 4 cents; he then went to another store and did the same, 
and then had 4 cents remaining; how much money had he at 
first? Ans. 4 cents. 

1G. A lady being asked her age, said that if her age were in- 
creased by its ^, the sum would equal 3 times her age 12 years 
ago; what was her age? Ans. 20. 

17. A lady being asked the hour of day, replied that f of the 
time past noon equaled \ of the time to midnight, minus \ of an 
hour; what was the time? Ans. 6 o'clock, P. M. 



162 FINKEL'S SOLUTION BOOK. 

18. What is the hour of day if -J- of the time to noon equals 
the time past midnight? Ans. 9 o'clock, A. M. 

19. A person being asked the time of day, said f of the time 
to midnight equals the time past midnight ; what was the time? 

Ans. 9 o'clock, A. M. 

20. A traveler on a train notices that 2J times the number of 
spaces between the telegraph poles that he passes in a minute is 
the rate of the train in miles per hour. How far are the poles 
apart? Ans. 198 feet. 

21. C's age at A's birth was 5-^ times B's age, and now is the 
sum of A's and B's ages, but if A were now 3 years younger and 
B 4 years older, A's age would be f of B's age. Find their ages. 

Ans. A's, 72 years; B's, 88 years; C's, 160 years. 

22. In the above problem change the last and to or, and what 
is their ages? Ans. A's, 36 ; B's, 44, and C's, 80. 

23. I have four casks, A, B, C, and D respectively. Find 
the capacity of each, if f of A fills B, f of B fills C, and C fills 
T 9 ^- of D; but A will fill C and D and 15 quarts remaining. 

Ans. A 35 gal., B 15, C 11±, and D 20. 

24. A man and a boy can do a certain work in 20 days : if 
the boy rests 5^ days it will take them 22^ days; in what time 
can each do it? Ans. The man, 36 da. ; the boy, 45 da. 

25. A can do a job of work in 40 days, B in 60 days; after 
both work 3 days, A leaves ; when must he return that the work 
may occupy but 30 days? Ans. 22-J- days. 

26. If 8 men or 15 boys plow a field in 15 days of 9^ hr., 
how many boys must assist 16 men to do the work in 5 days of 
10 hr. each? Ans. 12 boys. 

27. Bought 10 bu. of potatoes and 20 bu. of apples for $11 ; 
at another time 20 bu. of potatoes and 10 bu. of apples for $13 ; 
what did I pay for each per bu. ? 

Ans. Apples 30/, potatoes 50/. 

28. A farmer sold 17 bu. of barley and 13 bu. of wheat for 
$31.55, getting 35/ a bu. more for wheat than for the barley. 
Find the price of each per bu. 

Ans. Barley 90/, wheat $1.25. 

29. After losing f of my money I earned $12; I then spent f 
of what I had and found I had $36 less than I lost; how much 
money had I at first? Ans. $60. 

30. In a company of 87, the children are ■§■ of the women, and 
the women |- of the men; how many are there of each? 

Ans. 54 men, 24 women, and 9 children. 



PROBLEMS. 163 

31. If 4 horses or 6 cows can be kept 10 days on a ton of hay, 
how long will it last 2 horses and 12 cows? Ans. 4 days. 

32. A, B, and C buy 4 loaves of bread, A paying 5 cents, B 
8 cents, and C 11 cents. They eat 3 loaves and sell the fourth 
to D for 24 cents. Divide the 24 cents equitably. 

Ans. A 5 cents, B 8 cents, and C 11 cents. 

33. A and B are at opposite points of a field 135 rods in com- 
pass, and start to go around in the same direction, A at the rate 
of 11 rods in 2 minutes and B 17 rods in 3 minutes. In how 
many rounds will one overtake the other? Ans. B 17 rounds. 

34. If a piece of work can be finished in 45 days by 35 men 
and the men drop off 7 at a time every 15 days, how long will it 
be before the work is completed? Ans. 75 days. 

35. A watch which loses 5 min a day was set right at 12 M., 
July 24th. What will be the true time on the 30th, when the 
hands of that watch point to 12? Ans. 12:30-^ P - M - 

36. A seed is planted. Suppose at the end of 3 years it pro- 
duces a seed, and on each year thereafter each of which when 3 
years old produce a seed yearly. All the seeds produced, do 
likewise ; how many seeds will be produced in 21 years? 

Ans. 1872. 

37. The circumference of a circle is 390 rods. A, B, and C 
start to go around at the same time. A walks 7 rods per minute, 
B 13 rods per minute in the same direction ; C walks 19 rods per 
minute in the opposite direction. In how many minutes will 
they meet? Ans. 195 min. 

38. If 12 men can empty a cistern into which water is run- 
ning at a uniform rate, in 40 min., and 15 men can empty it in 30 
min., how long will it require 18 men to empty it? 

Ans. 24 men. 

39. Four men A, B, C, and D, agree to do a piece of work in 
130 days. A gets 42d., B 45d., C 48d., and D 51d., for every 
day they worked, and when they were paid each man has the 
same amount. How many days did each work? [da. 
Ans. A 35-IHHrf da -> B 33ffff da., C Slfffo da., and D 29ff|f 

40. A fountain has four receiving pipes, A, B, C, and D; A, 
B, and C will fill it in 6 hours; B, C, and D in 8 hours; C, D, 
and A in 10 hours; and D, A, and B in 12 hr.: it also has four dis- 
charging pipes, W, X, Y, and Z ; W, X, and Y will empty it in 
6 hours; X, Y, Z in 5 hours; Y, Z, and W in 4 hours ; and Z, 
W, and X in 3 hours. Suppose the pipes all open, and the 
fountain full, in what time will it be emptied? Ans. 6^ hours. 



164 



FINKEL'S SOLUTION BOOK. 



CHAPTER XVIII. 

ALLIGATION. 

1. Alligation is the process employed in the solution of 
problems relating to the compounding of articles of different 
values or qualities. 

2 Alii nation \ L A1 % ation Medial. 
a. JLll%gat%on j 2 Alligation Alternate. 

I. ALLIGATION MEDIAL. 

1. Alligation Medial is the process of finding the mean, 
or average, rate of a mixture composed of articles of different 
values or qualities, the quantity and rate of each being given. 

I. A grocer mixed 120 lb. of sugar at 5/ a pound, 150 lb. at 
6/., and 130 lb. at 10/.; what is the value of a pound of 
the mixture? 

1. 120 lb. @5/=$6.00, 

2. 150 lb. @6/=$9.00, and 
11.^3. 180 lb. @10?=$13.00. 

4. 400 lb. is worth $28.00. 
.5. .-. 1 lb. is worth $28-j-400=$.07=7 cents. 
III. .'. One pound of the mixture is worth 7 cents. 

(Stod. Comp. A., p. 244, prod. 8.) 



II. ALLIGATION ALTERNATE. 

1. Alligation Alternate is the process of finding in 
what ratio, one to another, articles of different rates of quality or 
value must be taken to compose a mixture of a given mean, or 
average, rate of quality or value. 

CASE I. 

Given the value of several ingredients, to make a compound of 
a given value. 

I. What relative quantities of tea, worth 25, 27, 30, 32, and 
45 cents per lb. must be taken for a mixture worth 28 
cents per lb. 



Diff. 



Bal. 



28/ 



Solution. — In average, 
the principle is, that the 
gains and loses are equal. 
We write the average price 
and the particular values 25, 
27? 30, 32, and 45 as in the margin. This is only a convenient 



f25/ 

27/ 


3/ 
1/ 


2 lb. 


17 lb. 


4 lb. 


19 1b. 
4 lb. 


30/ 

32/ 

U5/ 


2/ 

4/ 

17/ 


3 lb. 


3 lb. 


1 lb. 


3 lb. 
1 lb. 
3 lb. 



ALLIGATION. 165 

arrangement of the operation. Now one pound bought for 25/ 
and sold in a. mixture worth 28/ there is a gain of 28/ — 25/, or 
3/; one pound bought at 27/ and sold in a mixture worth 28/, 
there is a gain of 28/ — 27/, or 1/; one pound bought at 30/ and 
sold in a mixture worth 28/ there is a loss of 30/ — 28/, or 2/ ; 
one pound bought at 32/ and sold in a mixture worth 28/, there 
is a loss of 32/ — 28/, or 4/i and one pound bought at 45/ and 
sold in a mixture worth 28/ there is a loss of 45/ — 28/, or 17/. 
Since the gains and losses are equal, we must take the ingredi- 
ents composing this mixture in such a proportion as to make the 
gains and losses balance. We will first balance the 25/ tea and 
the 30/ tea. Since we gain 3/ a pound on the 25/ tea, and lose 
2/ on the 30/ tea, how many pounds of each must we take so 
that the gain and loss on these two kinds may be equal? Evi- 
dently, we should gain 6/ and lose 6/. To find this, we simply 
find the L. C. M. of 3 and 2. Now if we gain 3/ on one pound 
of the 25/ tea, to gain 6/,. we must take as many pounds as 3/ 
is contained in 6/, which are 2 lb. If we lose 2/ on one pound 
of the 30/ tea, to lose G/, we must take as many pounds as 2/ is 
contained in 6/, which are 3 lb. Next, balance the 25-cent tea 
and the 45-cent tea. The L. C. M. of 3/ and 17/ is 51/. Now 
if we gain 3/ on one pound of the 25-cent tea to gain 51/, we 
must take as many pounds as 3/ is contained in 51/ which are 
17 lb. If we lose 17/ on one pound of the 45-cent tea, to lose 
51/, we must take as many pounds as 17/ is contained in 51/ 
which are 3 lb. Next, balance the 27-cent tea and the 32-cent 
tea. The L. C. M. of 1/ and 4/ is 4/. If we gain 1/ on one 
pound of the 27-cent tea, to gain 4/, we must take as many 
pounds as 1/ is contained in 4/, which are 4 lb. If we lose 4/ 
on one pound of the 32-cent tea, it balances the gain on the 27- 
cent tea. Placing the number of pounds to be taken of each 
kind as shown above, and then adding horizontally, we have 19 
lb. at 25/, 4 lb. at 27/, 3 lb. at 30/, 1 lb. at 32/, and 3 lb. at 
45/. It is not necessary to balance them in any particular order. 
All that must be observed, is that all the ingredients be used in 
balancing. 

Note. — To prove the problem, use Alligation Medial. 



CASE II. 

To proportionate the parts, one or more of the quantities, but 
not the amount of the combination, being given. 

I. How many bushels of hops, worth respectively 50, 60, 
and 75/ per bushel, with 100 bushels at 40/ per bushel, will 
make a mixture worth 65/ a bushel? 



166 



FINKEL'S SOLUTION BOOK. 



Dif. 



Bal. 



A. 65/. 



40/ 

50/ 
60/ 

75/ 



25/ 


2bu. 






15? 




2 bu. 




5/ 






2 bu. 


10/ 


5bu. 


3 bu. 


1 bu. 




X50. 



100 bu. 

100 bu. 
100 bu. 
.450 bu. 



B. 65/. 





Dif. 


f 


Bal. 






40/ 
50/ 

60/ 
75/ 


25/ 

15/ 

5/ 

10/ 


2bu. 
5bu. 


2 bu. 
3bu. 


2bu. 
1 bu. 





X50= < 



100 bu. 
2bu. 

2 bu. 
-254 bu. 



Solution. — In this solution, we proceed as in Case I. In A, 
we obtain the relative amounts to be used of each kind, which is 
2 bu. at 40/, 2 bu. at 50/, 2 bu. at 60/, and 9 bu. at 75/. But 
we are to have 100 bu. of the first kind. Hence, we must multi- 
ply these results by 100-f-2, or 50. Doing this, we obtain 100 
bu. at 40/, 100 bu. at 50/, 100 bu. at 60/, and 450 bu. at 75/. 

Since either or both of the balancing columns, except the first, 
may be multiplied by any number whatever without affecting 
the average, it follows that there are an infinite number of re- 
sults satisfying the conditions of the problem. Since we are to 
have 100 bu. at 40/, the first column can be multiplied by 
only 50. 

In B, we have multiplied the first column by 50 and added in 
the results in the other two columns. This gives us 100 bu. at 
40/, 2 bu. at 50/, 2 bu. at 60/, and 254 bu. at 75/. The second 
and third columns may be multiplied by any number whatever. 
But the first must always must be multiplied by 50, because we 
are to have 100 bu. at 40 cents per bushel. 

(R. H. A., p. 338,prob. 2.) 

I. How much lead, specific gravity 11, with -J- oz. copper, 
sp. gr. 9, can be put on 12 oz. of cork, sp. gr. ^, so that the three 
will just float, that is, have a sp. gr. (1) the same as water? 



1 
1 1 


I 

I I 


r 

3 


r 


> 


i 


1 

9 


8 
9 




O 


>xi 


V 35 2 < 

A 2T 


4 


3 


10 
TI 


8 
9 

L. 










I 





'391 oz.=2 lb. 1\ 



oz. 



4 oz. 



112 oz. 



SoLUTiON.-^The specific gravity of any body is the ratio 
which shows how many times heavier the body is than an equal 



ALLIGATION, 167 

volume of water. Thus, when we say that the specific gravity 
of lead is 11, we mean that a cubic inch, a cubic foot, a cubic 
yard, or any quantity whatever is 11 times as heavy as an equal 
quantity of water. 

Now if a cubic inch (say) of lead be immersed in water, it 
will displace a cubic inch of water ; and since it weighs 11 times 
as much as a cubic inch of water, it displaces -Jy of its own 
weight. Hence, to have equal weights of water and lead we 
must take only -^ T as much lead as water. Now since a volume 
of water and TT as much lead have the same weight, and in 
the proper combination have a volume of 1, since the sp. gr. of 
the combination is 1, there is a loss of 1 — Jy, or TT , in volume on 
the part of the lead. For the same reason, there is a loss of -| in 
volume on the part of the copper, and 3 on the part of the cork. 
Balancing, we see that we must take 3 volumes of lead with Tr 
volumes of cork, a unit volume of water being the basis, in order 
that the two substances will just float, i. c, have a specific grav- 
ity (1). In like manner, we must take 3 volumes of copper 
with -| volumes of cork. Now since we must always take 3 vol- 
umes of lead for every TT volumes of cork, it is evident that the 
weights of the substances are in the same proportion. Hence, 
we may say, we must take 3 oz. of lead with every TT oz. ot 
cork, and 3 oz. of copper with every -§■ oz. of cork. 

But we are to have only \ oz. of copper. Hence, we must 
multiply the second balancing column by some number that will 
give us \ oz. of copper, i. c, we must multiply 3 by some number 
that will give us \. The number by which we must multiply is 
•J-j-3— -^. But multiplying •§ by jr, we get ^ 4 T oz. of cork. But 
we are to have altogether 12 oz. of cork. Hence we must yet 
have 12 oz. — 2 4 T oz.= 3 2 2 T ° oz. To produce this, we must multi- 
ply TT by some number that will give 3 2 2 f ° oz. This number is 
3 ? 2 7 p_j_i_o == 3 5 5_2 > But we must also multiply 3 by ¥ T °. This will 
give us 39^ oz.=2 lb. 7^ oz. of lead. Hence, we must use 2 lb. 7-g- 
oz. of lead, so that the three will just float. 

(R. H. A., p. SS9,prob. 7.) 

I. How many shares of stock, at 40%, must A buy, who has 
bought 1-20 shares, at 74%, 150 shares, at 68%, and 130 
shares, at 54%, so that he may sell the whole at 60%, 
and gain 20% ? 

'(1.) 100%=the average cost. 

(2.) 20%=gain. 

(3.) 120%=the average selling price. 

1 { (4.) 60%=the average selling price. 

(5.) .• 120%=60%. 

(6.) l^= T ^of 60%=-£%. 

(7.) 100%=100 times ^%=50%, the average cost. 



168 



FINKEL'S SOLUTION BOOK. 



J 



2. 



f(l-) 

(2.) 
(3.) 



120 shares 
150 shares 
130 shares 



74%= 8880%. 
68%=10200%. 
54%= 7020%. 



(4.) .*. 400 shares are worth 26100%, and 

(5.) 1 share is worth 26100 %-7-400=65i%, the average. 



3. 50% 



40 % 
65i% 



10 



151% 



154; shares. 
10 shares. 



{610 shares. 
400 shares. 



III. .-. He must take 610 shares. (R. H. A., p. 339, prob. 8.) 

Explanation. — Since 60% is the average selling price, and his gain is 20%, 
it is evident that his average cost is 60%-f-1.20, or 50%. In step 3, we find 
that the average cost of the 400 shares is 65|%. Hence, the problem is the 
same as to find how many shares at 40%, must A buy who has 400 shares at 
at an average of 65^% so that his average cost will be 50%. Balancing, 
we find that he must take 15£ shares at 40% with 10 shares at 65}%. But. 
he has 400 shares at 65£%. Hence, we must multiply the balancing col- 
umn by 400-i-lO, or 40. This gives 610 shares at 40%. 

CASE III. 

To proportion the parts, the amount of the whole combination 
being given. 

I. How many barrels of flour, at $8, and $8.50, with 300 
bbl. at $7.50, 800 bbl. at $7-80, and 400 bbl. at $7.65, 
will make 2000 bbl. at $7.85 a bbl.? 



II. 



III. 



300 bbl. @ $7.50 a bbl —$2250. 
800 bbl. @ $7.80 a bbl.=$6240. 
400 bbl. @ $7.65 a bbl. =$3060. 
are worth 



.-. 1500 bbl. are worth $11550. 

$7.85=the average price per bbl. of 2000 bbl. 

... $15700=2000 X$7.85=the value of 2000 bbl. 

... $15700— $11550=$4150=the value of 2000 bbl.— 1500 

bbl., or 500 bbl. 
... $8.30=$4150-r-500=the average value of 1 bbl. 



.30 



$8.00 
$8.50 



$.30 



2 bbl. 

3 bbl, 



X(500-f-5)= 



200 bbl. 



1. 200 bbl 

2. 300 bbl. at 



$.20 3 bblj 1300 bbl. . 

5 bbl. 
at $8.00 per bbl. must be taken with 



1.50 per bbl. 

(R. H. A., p. 339, prob. 2.) 



I. A dealer in stock can buy 100 animals for $400, at the fol- 
lowing rates: calves, $9; hogs, $2; lambs, $1; how many may 
he take of each kind ? 



ALLIGATION. 



169 



Bal. 



$1 



$9 




10117 
60'52 

30|31 



31J38 
3628 
33J34 



Explanation. — A lamb bought for $1 and sold for $4 is a gain of $3; a 
hog bought for $2 and sold for $4 is a gain of $2; and a calf bought for $9 
and sold for $4 is a loss of $5. We must make the gains and loses equal. 
The L. C. M. of $3 and $5 is $15. If we gain $3 on one lamb to gain $15 
we must take as many lambs as $3 is contained in $15, which are 5 lambs. 
If we lose $5 on one calf, to lose $15, we must take as many calves as $5 is 
contained in $15, which are 3 calves. The L. C. M. of $2 and $5 is $10. If 
we gain $2 on one hog, to gain $10, we must take as many hogs as $2 is 
contained in $10, which are 5 hogs. If we lose $5 on one calf, to lose $10, 
we must take as many calves as $5 is contained in $10, which are 2 calves. 
Adding the balancing columns, considering them as abstract numbers, we 
have 8 and 7- 8+7=15. 100-rl5=6f. •*• Multiplying each balancing 
column by 6|, will give 33£ lambs, 33£ hogs, and 33-J calves. But this result 
is not compatible with the nature of the problem. Hence we must see if 
we can take a number of 8's and a number of 7's that will make 100. By 
trial, we find that txuo 8's and twelve 7's will make 100. Hence, multiplying the 
first column by 2 and the second by 12, and adding the columns horizon- 
tally, we have for our result, 10 lambs, 60 hogs, and 30 calves. Again, we 
find, by trying three 8's, four 8's, and so on, that nine 8's taken from 100, 
will leave 28 which is four 7's. Hence, nine 8's and four 7's will make 100. 
Then, multiplying Ihe first column by 9 and the second by 4. and adding 
the columns horizontally, we have for a second result 45 lambs, 20 hogs, 
and 35 calves. Now these are the only answers that can be obtained by 
taking an integral number of 8's and integral number of 7's to make 100. 
But other answere may be obtained by taking 8 a fractional number of 
times, and 7 a fractional number of times to make 100. Suppose, for illus- 
tration, we try to take a number of thirds 8 times. We find that 8 taken 6- 
third times and 7 taken 36 third times will make 100. Multiplying the 
first column by % and the second by \ e , and adding the columns horizon- 
tally, we have, for a result, 10 lambs, 00 hogs, and 30 calves — the same as 
that obtained by taking 8 twice and 7 twelve times. Again, we find, that 
8 taken 13 third times and 7 taken 28-third times will make 100. Multiply- 
ing and adding as before we find that our results are fractional. Hence, we 
can not take a fraction whose denominator is three. It is clear that we 
must take a fraction whose denominator will reduce to unity w^hen being 
multiplied by 5. Hence, if we try to take 8 a number of fifths times and 7 
a number of fifths times to make 100, our results will all be integral. By 
trial, we find that 8 taken 3-fifths times and 7 taken 68-fifths times will 
make 100. Multiplying and adding as before, we have, for our results, 3 
lamb*, 68 hogs, and 29 calves. Again, we find that 8 taken 10-fifths times 
and 7 taken 60-fifths times, will make 100. Multiplying and adding as be- 
fore, we have, for results, 10 lambs, 60 hogs, and 30 calves. Again, by trial, 
we find that 8 taken 17-fifths and 7 taken 52-fifths times will make 100. 
Multiplying the first column by y and the second by R g 3 , and adding the col- 
umns horizontally, we have, for results, 17 lambs, 52 hogs, and 31 calves. 
Continuing the process, we find nine admissible answers. These are 
the only answers, satisfying the nature of the problem. 



70 



FINKEL'S SOLUTION BOOK. 



CHAPTER XIX. 

SYSTEMS OF NOTATION. 

1. A System of Notation is a method of expressing 
numbers by means of a series of powers of some fixed number 
called the Radix, or Base of the scale in which the different 
numbers are expressed. 

2. The Jtadioc of any system is the number of units of one 
order which makes one of the next higher. 



Names of Systems. 


Radix. 


Names of Systems. 


Radix. 


Unitary 

Binary 

Ternary 

Quaternary 

Quinary 

Senary 

Septenary 

Octonary - 


1 

2 
3 

4 
5 
6 

7 
8 


Nonary 

Decimal, or Denary 

Undenary 

Duodenary, 

Vigesimal, 

Trigesimal, 

Sexagesimal, 

Centesimal, 


9 
10 
11 
12 

20 

30 

60 

100 



4. In writing any number in a uniform scale, as many dis- 
tinct characters, or symbols, are required as there are units in the 
radix of the given system. Thus, in the decimal system, 10 char- 
acters are required; in the ternary, 3; viz., 1, 2, and 0; in the 
senary, 6; viz., 1, 2, 3, 4, 5, and 0; and so on. 

5. Let r be the radix of any system, then any number, N, 
may be expressed in the form, 

N=ar n -\-br n - l +cr n - 2 +dr n - z -\- .... +pr 2 + qr-\- s, in which 
the co-efficients a, b, c, . . . . , are each less than r. 

To express an integral number in a proposed scale : Divide 
the number by the radix, then the quotient by the radix, and so 
on; the successive remainders taken in order will be the successive 
digits beginning from units f lace. 



SYSTEMS OF NOTATION. 171 

I. Express the common number, 75432, in the senary system. 

1. 6 )75432 

2. 6 )12572+0 

3. 6 )2095+2 

4. 6)349+1 



IIJ 



5. 6 )58+1 

6. 6)9+4 



7. 1+3 

III. .'. 75432 in the decimal system=1341120 expressed in the 
senary system. 
I. Transform 3256 from a scale whose radix is 7, to a scale 
whose radix is 12. 



II. 



(1. 12 )3256 
2. 12)166+4 



3. 12)11+1 



4. 0+8 

III. .'. 3256 in the septenary system=814 in the duodenary 

system. 
Explanation. — In the senary system, 7 units of one order make one of the 
next higher. Hence, 3 units of the fourth order = 7X3, or 21, units of the 
third order. 21 units +2 units =23 units. 23+12 = 1, with a remainder 
11. 11 units of the third order =77 units of the second order. 77 units 
+5 units=82 units. 82+12=6) with a remainder 10. 10 units of the 
second order=70 units of the first order. 70 units +6 units=76 units. 76 
+12=6, with a remainder 4. Hence, the first quotient is 166, with a re- 
mainder 4. Treat this quotient in like manner, and so on, until a quo- 
tient is obtained, that is less than 12. 

I. What is the sum of 45324502 and 25405534, in the senary 
system? 

45324502 

25405534 

115134440 

Explanation. — 4+2=6. 6—6=1, with no remainder. Write the and 
carry the 1. 3 + 1=4. Write the 4. 5 + 5=10- 10+6=1, with a remainder 
4. Write the 4 and carrv the 1. 5-|-4-f-l=10. 10+6=1, with a remain- 
der 4. Write the 4 and carry the 1. 0+2+1=3. Write the 3. 4+3=7. 
7+6+1 with a remainder 1. Write 1 and carry 1 5+5 + 1=11. 11+6= 
1, with a remainder 5. Write the 5 and carry the 1. 2+4+1=7. 7+6= 
1, with a remainder 1. Write 1 and carry 1. The result is 115134440. 

I. What is the difference between 24502 and 5534 in the 
octonary system? 

24502 

5534 

16746 

Explanation. — 4 cannot be taken from 2. Hence, borrow one unit from 
a higher denomination. Then (2+8)— 6=4. (8—1)— 3=4. 5 from (4+8) 
—7. 5 from (3+8)=6. Hence, the result is 16746. 



172 FINKEL'S SOLUTION BOOK. 

I. Transform 3413 from the scale of 6 to the scale of 7. 
fl. 7 )3413 
II.i2. 7 )310+3 
[3. 7 )24+3 
2+2 

III. .-. 3413 in the senary system=2233 in the septenary sys- 
tem. 

I. Multiply 24305 by 34120 in the senary system. 

24305 
34120 

530140 
24305 
150032 
121323 



1411103040 

Explanation. — Multiplying 5 by 2 gives 10. 10+6=1, with a remainder 
4. Write 4 and carry 1 to the next order. 2 times 0=0. 0+1=1. Write 
the 1, 2 times 3=6. 6+6=1, with a remainder 0. Write the and 
carry the 1 to the next higher order. 2 times 4=8. 8+1=9. 9-4-6=1, 
with a remainder 3. Write 3 and carry the 1 to the next higher order. 
2 times 2=4. 4+1=5. Write 5. Multiply in like manner by 1, 4, and 3. 
Add the partial products, remembering that 6 units of one order, in the sen- 
ary system, uniformly make one of the next higher. 

I. Multiply 2483 by 589 in the undenary system, or system 
whose radix is 11. 

We must represent 10 by some character. Let it be t. 

2483 
589 



1/985 
1/502 
11184 
13322/5 

Explanation. — In the undenary system, 11 units of one order uniformly 
make one of the next higher order. 9 times 3=27. 27-*-ll=2, with a re- 
mainder 5. Write 5 and carry the 2 to the next higher order, or second or- 
der. 9 times 8=72. 72+2=74. 74-f-ll=6, with a remainder 8. Write 8 
and carry the 6 to the next higher order, or third order. 9 times 4=36. 
36+6=42. 42-4-11=3, with a remainder 9. Write 9 and carry the 3 to the 
next higher order, or the fourth order. 9 times 2=18. 18+3=21. 21 -+-11 
=1, with a remainder t. Write t and carry the 1 to the next higher order. 
Multiply in like manner by 8 and 5. Add the partial products, remember- 
ing that 11 units of one order equals one of the next higher. Wherever 
10 occurs, it must be represented by a single character t. 



SYSTEMS OF NOTATION. 173 

I. Divide 1184323 by 589 in the duodenary system. 

In the duodenary system, we must have 12 characters ; viz., 1, 
2, 3, 4, 5, 6, 7, 8, 9, /, e, and 0. t represents 10 and e, 11. 
589)1184323(2486 
e56 
22/3 
1M) 
3*32 
39/0 



1523 
1523 



Explanation. — In the duodenary system, 12 units of one order make one 
of the next higher. 1184 will contain 589, 2 times. Then multiply the 
divisor, 589, by 2 thus: 2 times 9=18. 18-j-12=1, with a remainder G. 
Write the 6 and carry the 1. 2 times 8=16. 16+1=17. 17-j-12=1, with a 
remainder 5. Write the 5 and carry the 1. 2 times 5=/". /+l=e. 
Write the e. Then subtract. 6 from (12+4)=/, 5 from 7=2, and e from (12 
+D=2. 

Hence, the first partial dividend is 22 /. Bring down 3. Then 22t3 will 
contain 589, 4 times. Multiply as before. By continuing the operation we 
obtain 2483 for a quotient. 

I. Divide 95088918 by #4, in the duodenary system. 

//4)95088918(/4/^ 
9074 
4548 
3754 




/*58 
/c58 



I. Extract the square root of 11122441 in the senary system. 

lil22441(2405 
2X2= 4 



44 



312 
304 



2x24=52 
2X240=520 5 



42441 
42441 



Explanation. — The greatest square in 11 expressed in the senary system 
is 4. Subtracting and bringing down the next period, we have 312 for the 
next partial dividend. Doubling the root already found and finding how 
many times it is contained in 312 expressed in the senary system, we find it 
is 4. Continuing the process the same as in the decimal system, the re- 
sult is 2405. 



174 



FINKEL'S SOLUTION BOOK, 



Extract the square root of 11000000100001 in the binary 
system. 



11000000100001(1101111 
1 



101 

11001 

110101 

1101101 

11011101 



1000 
101 



110000 
11001 



1011110 
110101 



10100100 
1101101 



11011101 
11011101 



III. 
I. 

II. 
III. 



( Todhunter' s Algebra, p. 255, Ex. 23.) 
Find in what scale, or system, 95 is denoted by 137. 

1. Let r=the radix of the system. Then 

2 . r»+3r+7=95, 

3. r*+3r=95— 7=88, and 

4. r 2 4-3r-|"f==88+f = 3 f\ by completing the square. 

5. r-j-f= 1 -g 9 , by extracting the square root, and 

6. f=V — 1== 1 ^=8, the radix of the system. 

.". 95 is denoted by 137 in the octonary system. 

( Todhunter 's Alg., p. 255, prob. 26.) 

Find in what system 1331 is denoted by 1000. 

1. Let r=the radix of the system. Then 
r 4_|_0r 8 +0r 8 +0r+Q=1331, or 
r 4 =1331. Whence 
r=\/i33i =11, the radix of the system. 

1331 is denoted by 1000 in the undenary system. 

( Todhunter" s Alg., -p. 255, prob. 28.) 



MENSURATION. 175 

CHAPTER XX. 

MENSURATION 

1. Mensuration is that branch of applied mathematics 
which treats of geometi'ical magnitudes . 

2. Geometrical Magnitudes arc lines, surfaces, and 
solids. 

3. Geometrical Magnitudes. 



C. Solid. B. Surface. A. Lira 



0< 4- CO to ►-» tO h-i to h- 

o*2.' 2 ^ 2. c s* a :? 

* ~ D ■— v >-t Z, "S m 

c» n> S* 4. < S ^ -• 

M • rt s* rti P> 0> TO 

p ■ S* S. P* e-s: 



lO -- to r- 1 Oi 4- CO 10 (-» ^_ ^ _ .. Oot-H^M~. 

0^0^. P - ^' 2, 
0*0*^ C/? J 'sag ^oj< 



►o i»a 



3" P o> - S 3 3 ~oS *3 



3* o 

a & & £ K. a K 



a 



1 3 Co 

o 



sT nP 2t 



pa 3 

TO 2. TO 

o s- r 



H » -• > ^ -• > ? 



°:<w ;; 



p 3 Cfl " rt 



!^^a-o & *° 



3" 



n> . ■— n» 



p ? 
en 

f& o 



<T> ft rT 

a • So o so 



lO <-> Oi 4- CO tO i- 1 



P 

pen. ^•p rt < 2. v, _^_ 

» P ~ 4. X 3 S3 p / v I V 

Q- 3 ,D J P £" O- 3 lO M CO lO M to M 

2? §— o°§ S^ ■* ■*.*.*.-..*.* 

E.." ° r-£Ler< PoSK-TO-gSogo 

(TO 3 « g = ff P <J> P 2 



c 






^ ~ fD 

°~ = P 5 _ ^ 

C ?3 p 

S? ^ 

° 3 

3 crq 

a = 



176 FINKEL'S SOLUTION BOOK. 

4. A Line is a geometrical magnitude having length, with- 
out breadth or thickness. 

5. A Straight Line is a line which pierces space evenly, 
so that a piece of space from along one side of it will fit any side 
of any other portion. 

6. A Curved Lifie is a line no part of which is straight. 

7. A Surface is the common boundary of two parts of a 
solid, or of a solid and the remainder of space. 

8. A Plane Surface, or Plane, is a surface which di- 
vides space evenly, so that a piece of space from along one side 
of it will fit either side of any other portion of it. 

9. A Curved Surface is a surface no part of which is 
plane. 

10. A Polygon (TLokvyoDvoS, from IIoXvs, many, and 
yoovia, angle) is a portion of a plane bounded by straight lines. 

11. A Circle (xipuos, circle, ring) is a portion of a plane 
bounded by a curved line every point of which is equally distant 
from a point within called the center, 

12. An Ellipse ( eWeiipiS) is a portion of a plane bounded 
by a curved line any point from which, if two straight lines are 
drawn to two points within, called the foci, the sum of the two 
lines will be constant. 

13. A Triangle (Lat. Triangulum, from tries, trid, 
three, and angulus, corner, angle) is a polygon bounded by three 
straight lines. 

14. An Angle is the opening between two lines which 
meet in a point. 

II. Straight Angle. 

2. Right Angle. 

o nw \ !• Acute - 

3. Oblique j 2 0btuge 

16. A Straight Angle has its sides in the same line, and 
on different sides of the point of meeting, or vertex. 

17. A Hight Angle is half of a Strait Angle, and is 
formed by one straight line meeting another so as to make the 
adjacent angles equal. 

18. An Oblique Angle is formed by one line meeting 
another so as to make the adjacent angles unequal. 

19. An Acute Angle is an angle less than a right angle. 

20. An Obtuse Angle is an angle greater than a right 
angle. 



MENSURATION. 177 

/ 

21. A Might Triangle is a triangle, one of whose angles 
is a right angle. 

22. An Oblique- Angled Triangle is one whose angles 
are all oblique. 

23. An Isosceles Triangle is one which has two of its 

sides equal. 

24. A Scalene Triangle is one which has no two of its 

sides equal. 

25. An Equilateral Triangle is one which has all 
the sides equal. 

26. A Quadrilateral (Lat. quadrilaterus , from quatuor, 
four, and latus, lateris, a side) is a polygon bounded by four 
straight lines. 

27. A Parallelogram (napaXkrjXoypafjL)j.oy, from 
HapaWrjXoS, parallel, and ypajujurj , a stroke in writing, a line) 
is a quadrilateral having its opposite sides parallel, two and two. 

28. A Might Parallelogram is a parallelogram whose 
angles are all right angles. 

29. An Oblique Parallelogram is a parallelogram 
whose angles are oblique. 

30. A Mectangle (Lat. rectus, right, and angulus, an 
angle) is a right parallelogram. 

31. A Square is an equilateral rectangle. 

32. A Mhomboid (pojA/3osidr/S, from po/xfios, rhomb, and 
ei'dos, shape) is a parallelogram whose angles are oblique. 

33. A Mhombus (popiflos, from ptjufieiv, to turn or whirl 
round) is an equilateral rhomboid. 

34. A Pentagon (nevTayoovov, TJevrs, five, and yoovia, 
angle) is a polygon bounded by five sides. Polygons are named 
in reference to the number of sides that bound them. A Hexa- 
gon has six sides; Heptagon, seven; Octago?i, eight; JVo?iagon, 
nine; Decagon, ten; Undecagon, eleven; Dodecagon, twelve; 
Tridecagon, thirteen ; Vctradccagon , fourteen; Pentedecagon, 
fifteen; Hcxdccago?i, sixteen; Heptadecagon, seventeen; Octa- 
decagon, 'eighteen; En?ieadccagon y nineteen; Icosagon, twenty; 
Icosaisagon, twenty-one ; Icosadoagon, twenty-two; Icosatriagon, 
twenty- three ; Icosatetragon, twenty -four ; Icosapentegon, twenty- 
five; Icosahcxagon, twenty-six; Icos a kept agon, twenty-seven; 
Icosaoctago7i, twenty -eight ; Icosacnneagon, twenty-nine; Tria- 
contagon, thirty; Tricontaisagon, thirty- one; Tricontadoagon, 
thirty-two; Tricontatriago)i. thirty-three; and so on to Tessa- 
racontagon, forty; Peutecontagon, fifty; Hcxacontagon, sixty; 



178 .. FINKEL'S SOLUTION BOOK. 

Hebdomacontagon , seventy; Ogdoacontagon, eighty; Encnacon- 
tagon, ninety; Hecatonagon , one hundred; Diacosiagon, two 
hundred; Triacosiagon, three hundred; Tetracosiagon, lour hun- 
dred; Pentecosiagou, five hundred; Hexacosiagon, six hundred; 
Heptacosiagon, seven hundred ; Oktacosiagon, eight hundred ; 
Enacosiagou, nine hundred; Ckiliagon, one thousand; &c. 

35., ^4 Spherical Surface is the boundary between a 
sphere and outer space. 

36. A Conical Surface is the boundary between a cone 
and outer space. 

37. A Cylindrical Surface is the boundary between 
the cylinder and outer space. 

38. A Solid is a part of space occupied by a physical body, 
or marked out in any other way. 

39. A Polyhedron (IloXvedpoS, from IIoXvs, many, and 
sdpa, seat, base) is a solid bounded by polygons. 

40. A Prism is a polyhedron in which two of the faces 
are polygons equal in all respects and having their homologous 
sides parallel. 

41. The Altitude of a prism is the perpendicular distance 
between the planes of its bases. 

42. A Triangular Prism is one whose bases are trian- 
gles. 

43. A Quadrangular Prism is one whose bases are 
quadrilaterals. 

44. A Parallelopipedon is a prism whose bases are 
parallelograms. 

45. A Might Parallelopipedon is one whose lateral 
edges are perpendicular to the planes of the bases. 

46. A Rectangular Parallelopipedon is one whose 

faces are all rectangles. 

47. A Cube (xvfios, a cube, a cubical die) is a rectangular 
parallelopipedon whose faces are squares. 

48. A Might Prism is one whose lateral edges are per- 
pendicular to the planes of the bases. 

49. An Oblique Prism is one whose lateral edges are 
oblique to the planes of the bases. 

50. A Pyramid (IIv pa juidos) is a polyhedron bounded 
by a polygon called the base, and by triangles meeting at a com- 
mon point called the vertex of the pyramid. 



MENSURATION. 179 

51. The Convex Surface of a pyramid is the sum of the 

triangles which bound it. 

52. A Might Pyramid is one whose base is a regular 
polygon, and in which the perpendicular, drawn, from the vertex 
to the plane of the base, passes through the center of the base. 
The perpendicular is called the axis. 

53. A Tetrahedron (rHpa, four, and eSpa, seat, base) 
is a pyramid whose faces are all equilateral triangles. 

54. The Altitude (Lat. Altitudo, from alt us, high, and 
ude denoting state Or condition) of a pyramid is the perpendicu- 
lar distance from the vertex to the plane of the base. 

55. The Slant HeigJ^t of a pyramid, is the perpendicu- 
lar distance from the vertex to any side of the base. 

56. A Triangular Pyramid is one whose base is a 
triangle. 

57. An Octahedron (onraedpos, from our a, eight, and 
edpa, seat, base) is a polyhedron bounded by eight equal equi- 
lateral triangles. 

58. A Dodecahedron (8 cod eua, twelve, and i'dpa, seat, 
base) is a polyhedron bounded by twelve equal and regular 
pentagons. 

59. An Icosahedron (i'nwai, twenty, and e$pa, seat, base) 
is a polyhedron bounded by twenty equal equilateral triangles. 

60. A Cylinder (uvXivdpos, from, jiv\ivdpeiv, uvXsiv, to 
roll) is a solid bounded by a surface generated by a line so mov- 
ing that every two of its positions are parallel, and two parallel 
planes. 

61. The Axis (aHoov) of a cylinder is the line joining the 
centers of its bases. 

62. A Might Cglhldcr is one whose axis is perpendicu- 
lar, to the planes of the bases. 

63. A Cone (hgdvos, from Skr. co, to bring to a point) is 
a solid bounded by a surface generated by a straight line moving, 
so as always to pass through a fixed point called the apex,, and a 
plane. 

64. A Might Cone is a solid generated by revolving a 
right-angled triangle about one perpendicular. 

65. An Oblique Cone is one in which the line, called 
the axis, drawn from the apex to the center of the base is not 
perpendicular. 

66. The FvU8tUfofa.(h%t.frustuni, piece, bit) of a pyra- 
mid or a cone is the portion included between the base and a 
parallel section. 

67. A Sphere (Gcpalpa) is a solid bounded by a curved 



180 



FINKEL'S SOLUTION BOOK. 



surface, every point of which is equally distant from a point 
within, called the center. 

Before we enter into the solution of problems in Mensuration, 
it will be necessary first to explain a difficulty which we en- 
counter. 

The common way of teaching that feet multiplied by feet give 
square feet is wrong ; for there is no rule in mathematics justify- 
ing the multiplication of one denominate number by another. 
If it is correct to say feet multiplied by feet give square feet, we 
might, with equal propriety, say dollars multiplied by dollars 
give square dollars — a product wholly unintelligible. In all our 
reasoning, we deal with abstract numbers alone or the symbols 
of abstract numbers. These do not represent lines, surfaces, or 
solids, but the relations between these numbers may represent 
the relations between the magnitudes under consideration. 

Suppose, for example, that the line AB contains 5 units, and 
the the line B C 4 units. Let a denote the abstract number 5, 
and b the abstract number 4. Then 
ab=20. Now this product ab is not 
a surface, nor the representation of 
a surface. It is simply the abstract 
number 20. But this number is ex- 
actly the same as the number of 
square units contained in the rectan- 
gle whose .sides are AB and B C, as 
may be seen by constructing the rec- 
tangle AB CD. Hence the surface of 
the rectangle is measured by 20 squares described on the unit of 
length. 

This relation is universal, and we may always pass from the 
abstract thus obtained by the product of any two letters, to the 
measure of the corresponding rectangle by simply considering 
the abstract units as so many concrete or denominate units. 

In like manner, the product of three letters abc is not a solid 
obtained by multiplying lines together, which is an impossible 
operation. It is simply the product of three abstract numbers 
represented by the letters a, b, and c, and is consequently an 
abstract number. But this number contains precisely as many 
units as there are solid units in the parallelopipedon whose edges 
correspond to the lines a, b, and c\ hence, we may easily pas's 
from the abstract to the concrete. Hence, if we wish to find the 
area of a rectangle whose width is 4 feet and length 6 feet, we 
simply say, 6x4=24 square feet. We pass at once from the ab- 
stract in the first member to the concrete in the second. 

It is a question whether pupils should be taught a falsehood 
in order that they may learn n truth. 

(See Bledsoe's Philosophy of Mathematics, pp. 97-106.) 



MENSURATION. 181 



I. PARALLELOGRAMS. 

Prob. I. To find the area of a parallelogram; whether it 
be a square, a rectangle, a rhomboid, or a rhombus. 

Formula. — A = lxb, where ^4=area, /=length, and b= 
breadth ; or, A=bX<*, where ^4=area, £=base, and «=altitude. 

Rule — Multiply the length by the breadth; or, the base by the 
altitude. 

I. What is the area of a parallelogram whose length is 15 
feet and breadth 7 feet? 

By formula, ^=/X^=lengthXbreadth=15X 7=105 sq. 
feet. 



II. 



1. 15 feet=length. 

2. 7 feet=breadth. 

3. .-. 15X7=105 sq. ft. 
^ =area. 

III. .-. The area i s 105 sq. 
ft. 




FIG. 4. 



Note. — The base is not necessarily the side toward the ground. Thus in 
the parallelogram ABCD,BC may be considered the base, in which case, 
the altitude would be the perpendicular distance BF, between the sides BC 
and AD. If HG and BC were given, we could not find the area of the par- 
allelogram because we have not the base and altitude given. 

I. What is the area of the parallelogram ABCD, if BC is 
26 feet and EFbO feet? 

By formula, A=aXo=BBx^ C=oOX 26=1300 sq. ft. 

II. 26 feet=j5C=base. 
2. 50 feet=i£./z==altitude. 
3. .'. 26X50=1300 sq. ft.=area. 
III. .-. The area of ABCZ)=1Z00 sq. ft. 

I. A floor containing 132 square feet, is 11 feet wide ; what 
is its length? 

By formula, A=lXo. •'. /=^4-t-£= 132-5-1 1=12 ft. 

II. 132 sq. ft.=area. 
2. 11 ft.=breadth. 
3. 132-7-11=12 ft.=length. 

III. .-. The floor is 12 ft. long. 



182 



FINKEL'S SOLUTION BOOK. 



Prob. II. The diagonal of a square being- given, to find 
tlie area. 

Formula. — A=d 2 -r-2. 

Rule. — Divide the square of the diagonal by 2, and the 
quotient will be the area. 

I. What is the area of a square 
whose diagonal is 8 chains? 

By formula, A=d M-2== 8 2 -f- 
2=32 sq. chains. 

8 ch.=length of diagonal— 
BD. 

64 sq. ch.=8xS=^ GFH= 
square described on the di- 
agonal BD. 

32 sq. ch.=64 sq. ch.-f-2= 

area of the square ABCD. 

32 sq. ch.=the area of the 



IIJ 




III. 
square 



FIG. 5. 



Prob. III. The area of a square being given, to find its 



II 



Formula. — d=x r 2A. 

Rule. — Extract the square root of double the area. 
I. The area of a square is 578 sq.ft. ; what is the diagonal? 
By formula, ^ r ==V'^4= : V / 2X578==\/Tl56=34 feet. 

1. 578 sq. ft.=area of the square. 

2. 1156 sq. ft.=2x578 sq. ft.=double the area. 
[3. 34 feet=\/ii5(3=the diagonal. 

III. .-. The diagonal is 34 feet. 

Prob. IT. The diagonal of a square being- given, to find 
its side. 

Formula. — S=y/\d*. 

Rule. — Extract the square root of one -half the square of the 
diagonal. 

I. What is the side of a square whose diagonal is 12 feet? 
By formula, 6 , =v / -^=V / p<T^=V / 72==6V'2=8.4852+ft. 

1. 12 ft.=the diagonal. 

2. 144 sq. ft.— 12 2 =square described on the diagonal. 

3. 72 sq. ft.=area of square whose side is required. 
.4. .-. 8.4852 ft— 6V2=V72= side of the square. 

.*. The side of the square is 8.4852-[-ft. 



II.- 



III. 



MENSURATION. 183 

Prob. V. To find the side of a square having- its area 
given. 

Formula.— S=\/a . 

Rule, — Extract the square root of the number denoting its 
area. 

I. What is the side of a square field whose area is 2500 
square rods? 

By formula, S=\/a=\/ '2500=50 rods. 

TT { 1. 2500 sq. rd.=area of the field. 

(2. 50 r d . = V 2500= side of the square field. 

III. .'. The side of the field is 50 rods. 

II. TRIANGLES, 

Prob. VI. Given the base and altitude of a right-angled 
triangle, to find the hypothenuse. 

Formula. — h=\Z«*+b*. 

Rule. — To the square of the base add the square of the alti- 
tude and extract the square root of the su?n. 

I. In the right angled triangle A CB , the base A C=~y() and 
the altitude Z?C=33 ; what is the hypothenuse? 

By formula, // == v V+Z>*=\/33 2 +50 2 =v / 1089+313G=V /: ^ 
=65. 

1. 56=^4 C=the base. 

2. 3136=56 2 =the square of the 
base. 

3. 33=i?C=the altitude. 

4. 10S9=33 2 =the square of the 
altitude. 

5. 4225=3 136+1089=the sum of »■ *• 
the squares of the base and altitude. 

6. 65=\/ 4225= the square root of the sum of the squares of 
the base and altitude=the hypothenuse. 

III. .*. The hypothenuse=65. 

Prob. VII. To find a side, when the hypot lien use and the 
other side are given. 



II. 




\ a=\Z/,2—i,-i. 
( b==y/hi—a*. 

Rule. — From the square of the hypothenuse subtract the square 
of the given side and extract the square root of the remainder. 



184 FINItEL'S SOLUTION BOOK. 

I. The hypothenuse of a right-angled triangle is 109, and 
the altitude 60; what is the base? 

By formula, b=yf h*— « 2 =\/l09 2 — 60 2 =V / 8281=91. 

1. 109=hypothenuse. 

2. 11881=109 2 =square of the hypothenuse. 

3. 60=the altitude. 
[I { 4. .3600=60 2 =the square of the altitude. 

5. 8281=11881— 3600=difference of the squares of the 
hypothenuse and altitude. 

6. 9 1= V8281=the square root of this difference=the base. 

III. .\ The base is 91. 

Remark. — When a=b, Ji—V 2a 2 =av 2. From this, we see that the diag- 
onal of a square is v 2 times its side. 

Prob. VIII. To find the area of a triangle, having- given 
the base and the altitude. 

Formula. — A=\a x b. 

Rule. — Multiply the base by the altitude and take half the 
product. 

I. What is the area of a triangle whose base is 24 feet and 
altitude 16 feet? 

By formula, A=%aX&=i X 16x24=192 sq. ft. 

1. 24 ft.=base. 

2. 16 ft.=altitude. 

3. 384 sq. ft.=16 X24 = pro- 
II. <J duct of base and altitude. 

4. 192 sq. ft.=| of 384 sq. ft.= 
half the product of the base 
and the altitude=area. 

III. .'. The area of the triangle is r . 

192 sq. ft. 

Prob. IX. To find the area of a triangle, having- g-iven 
its three sides. 

Forinula. — A=y/ S ( s—a ) ( s—b ) ( s—c ) , where s—% ( a -\-b-\-c). 

*Rllle. — Add the three sides together and take half the sum; 
from the half sum, subtract each side seperately; multiply the half 
sum and the three remainders together and extract the square root 
of the product. 

* Demonstration. — In Fig. 7, let A Cz=b, BC=a, and AB=c. In the 
right-angled triangle ADB, B D^—AB^—AD*, and in the right-angled 
triangle CZ>B, B£> 2 —BC 2 —Z>C 2 . .'. AB 2 —AD 2 =BC 2 —DC 2 . or c 2 — 




II. 



MENSURATION. 185 

I. What is the area of a triangle whose sides are 13, 14, and 
15, feet respectively? 

By formula, A=y/ 7{s^-a)(s— b)(s — c) = SJ21X (21—13) X (21— 
14)X(21— 15)=\/21X8X7X6 = V7056=84 sq. ft. 

'l. 42 ft.=13 ft.+14 ft.+15 ft.=sum of the three sides. 

2. 21 ft.=^ of 42 ft.==half the sum of the three sides. 

3. 21 ft.— 13 ft.=8 ft.=first remainder. 

4. 21 ft. — 14 ft.=7 ft.=second remainder. 

5. 21 ft. — 15 ft.=6 ft.=third remainder. [mainders. 

6. 7056=21 X6X7X8=product of half sum and three re- 

7. 84 sq. ft.= ^7056=square root of the product of the half 
sum and three remainders=the area of the triangle. 

III. .-. The area of the triangle is 84 sq. ft. 

Prob. X. To find the radius of a circle inscribed in a 
triangle. 

Formula.— R=2A^-( a+b+c ). 
*Rule. — Divide twice the area of the triangle by the sum of 
the three sides. 

I. Find the radius of a circle inscribed in a triangle whose 
sides are 3, 4, and 5 feet, respectively. 

"1. 6 sq. ft.= tis(s — a)(s — b)(s — c)=area of the triangle, 
by formula, Prob. IX. 

2. 12 sq. ft.=twice the area of the triangle. 

3. 12 ft.=3 ft.-|-4 ft.+5 ft.=sum of the three sides. 

4. .-. 1 ft.=12-r-12=twice the area divided by the sum of 
the sides=the radius of the inscribed circle. 

III. .-. The radius of the inscribed circle is 1 ft. 

AD 2 =<i 2 — DC 2 , whence c 2 —a 2 =AD 2 —DC 2 . But AD 2 — DC 2 — {AD 
+DC'(AD—DC)=b(AD—DC). .-. b(AD— DC)~c 2 — a 2 , and AD— DC 
=(r 2 — a 2 )-i-b. But AD-{-DC=b. .'. By adding the last two equations, we 

have 2AD= - L \-b= =— ^ ! whence AD— —L Since 

b b 2b 

BD 2 =AB 2 —AD 2 =c 2 —ylD 2 , if we substitute the value of AD just found, 

(c 2 — « 2 +b 2 \ 2 4b 2 c 2 —(c 2 —a 2 -\-b 2 ) 2 
wehave^ 2 =r 2 -[ ^-J = ,, 2 ^ ; = 

( 2 Z,r-f-r 2 — <t 2 -\-b 2 )(2bc— c 2 +a 2 — b 2 ) _ {b 2 -\- 2bc+c 2 — a 2 )[a 2 — (b 2 — 2bc+c 2 )] _ 
Ab 2 ~ Ab 2 — 

[(b+c) 2 -a 2 ]{a 2 —(b—c) 2 ] 

-i __ 9 

.-. BD=-^sl[{b+c) 2 — a 2 ]{a 2 — (b-c) 2 ] — pf s (s— a)(s— b)(s— c). Now the 

area of ABC—\A CXBD. 

... A=\b^BD=\byhj[{b+c) 2 -a 2 }[a 2 -{b-c) 2 ] 



II. 



=W [(H-Q 2 — « 2 ][tf 2 — (* — C ) Z ]= iV(H-^+«)(H- f — «){«+b— c){a— b+c). 
— ^(s—aXs—bKs—c), where 2s=(a+b+c). Q. E. D. 
*Note. — For Demonstration, see any geometry. 



186 FINKEL'S SOLUTION BOOK. 

Prob. XI. To find the radius of a circle, circumscribed 
about a triangle whose sides are given. 

_ _ abc abc 

Formula. — R= — = — , — ^ =- 

4A 4 */s(s—a)(s—6)(s—c). 

*Rllle. — Divide the ■product of the three sides by four times the 
area of the triangle. 

I. What is the radius of a circle circumscribed about a tri- 
angle whose sides are 13, 14, and 15 feet, respectively? 

1. 2730 cu. ft.=13Xl4Xi5=the product of the three sides. 

2. 84 sq. ft.= \s(s — a)(s — b)(s — c)=the area of the tri- 
• angle, by Prob. IX. [angle. 

II.<I 3. 336 sq. ft.=4x84 sq. ft.=four times the area of the tri- 
4. §1 ft.=2730-r-336=the product of the three sides divid- 
ed by four times the area of the triangle=the radius 
of the circumscribed circle 

III. .-. The radius of the circumscribed circle is 8^ ft. 

Prob. XII. To find the area of an equilateral triangle, 
having* given the side. 

Formula. — ^4=i V3i" 2 , where .?=side. This is what Prob. 
IX. becomes, when a=b=c. 

Rule. — Multiply the square of a side by \ <y/3,=.433013-f-. 

I. What is the area of an equilateral triangle whose sides 
are 20 feet? 

* Demonstration. — Let ABC be any triangle, and ABCE the circumscrib- 
ed circle. Draw the diameter BE, and 
draw.EC. Draw the altitude BD of the 
triangle ABC. The triangles ADB and 
BCE are similar, because both are right- 
angled triangles, and the angle BAD=zthe 
angle BEC. Hence, AB'-EB'-'-BD'. BC. 
Hence, ABXBC=BEXBD or ac=2RX 
BD. But, in the demonstration of Prob. 

2 

IX., we found BD=—s/ S (s—a)(s—b)(s—c). 

2 

.-. «c = 2i?X-rv/«(«-o)M)(«-«)- Whence 

abc abc 

R ~~ W^a) («-») i.9-c)~~4A ' FIG. 8* 



HH 



XL- 



MENSURATION. 187 

By formula, A=\ V3X20 2 =100 V3=173.205+sq. ft. 

1. 20 ft.=length of a side. 

2. 400 sq. ft.=20 2 =square of a side. 

3. 173.205 sq. ft.=4 V3X400=.433013X40O=jV3 times 
I the square of a side,=the area of the triangle. 

III. .'. The area of the equilateral triangle is 173.2054-sq. ft. 

Prob. XI JT. Tbe area and base of a triangle being* given, 
to cut oil a triangle containing- a given area, by a line run : 
ning- parallel to one of its sides. 

Formula, — ^'— ^^J^T' wnere A=area of the given 

triangle; £, the base of the given triangle; and A , the 
area of the portion to be cut off. 

Rule. — As the area of the given triangle is to the area of the 
triangle to be cut off, so is the square of the given base to the 
square of the required base. The square root of the result vjill 
be the base of the required triangle. 

I. The area of the triangle ABC is 250 square chains and 
the base AB, 20 chains ; what is the base of the trian- 
gle, area equal to 60 sq. chains, cut off by ED parallel 
to BC? 

By formula, AD=b=b^~=20^\^ ) =4 s T ! > =-9.7979 + ch. 

T. 250 sq. ch.=area of the given triangle ABC. 

2. 60 sq. ch.=area of the 
triangle AED. 

3. 20 ch.=base of the trian- 
<r\e ABC. 

4. .-. 250 sq. ch. : 60 sq. ch. 
:: 20 2 : AD 2 . Whence 

5. AD*= (400X60) —250 
=96. 

.6. .'. .4Z)= v ^=9.7979-f-ch. FIG. 9. 

III. .-. The base A£>==®.797$+ch. 

III. TRAPEZOID. 

Prob. XTY. To find the area of a trapezoid, having: given 
the parallel sides and the altitude. 

Formula, — A=\(b-\-b')a, where b and b' are the parallel 
sides and a, the altitude. 

Rule. — Multiply half the sum of .the parallel sides by the al- 
titude. 



II. 




188 



FINI'EL'S SOLUTION BOOK. 



II. 



4. 



15. 




I. What is the area of a trapezoid whose parallel sides are 
15 meters and 7 meters and altitude 6 meters? 

By formula, A=\(b+b' ) X^=|(15+7) X6=66 m 2 . 

1. 7 m.=Z>C,the length of one 
of the parallel sides, and 

2. 15 m.=AB, the length of 
the other side. 

3. 22 m.==7 m.-j-15 m.=sum 
of the parallel sides. 

11 m.=4 of 22 m — half 

the sum of the parallel 

sides. fIG. 10. 

66 m 2 .=6Xll=area of the trapezoid, ABCD. 

III. .*. The area of the trapezoid is 66 m 2 . 

IV. TRAPEZIUM AND IRREGULAR POLYGONS. 

Prob. XV. To find the area of a trapezium or any irreg- 
ular polygon. 

Rule. — Divide the figure into triangles, find the area of the 
triangles and take their sum. 

I. What is the area of the trapezium ABCD, whose diag- 
onal A C is 84 feet, and the perpendiculars DE and 
BF, 56 and 22 feet, respectively? 

1. 84 ft.=^4 C =kase of the 
triangle ADC. 

2. 56 ft.=/)^ = altitude ot 
ADC. 

3. /. 2352 sq. ft.=4(^4CX 
D^)=area of the trian- 
gle AD C. 

l.<!4. 84 ft.=AC=base of the 
triangle ABC. 

5. 22 {t=BB= altitude of 
ABC. 

6. .•.924sq.ft.=4(^4Cx^F) 
=area of the triangle ABC. 

7. 3276 sq. ft.=2352 sq. ft.+924 sq. ft— ADC+ABC= 
area of the trapezium ABCD. 

III. .-. The area of the trapezium ABCD is 3276 sq. ft. 

V. REGULAR POLYGONS. 

Prob. XVI. To find the area of a regrular polyg-on. 

Formula. — A—laXfl, where f is the perimeter and a, 
the apothem. 

Rule* — Multiply the perimeter by half the apothem. 




MENSURATION. 



189 



T^he Perimeter of any polygon is the sum of all its sides. 

The Apothem is the perpendicular drawn from the center to 
any side of the polygon. 

I. What is the area of a regular heptagon whose side is 
19.38 and apothem 20? 

By formula, A=^aXp=iX20X (7 X 19.38 )=1356.6. 
(1. 19.38=length of one side. 

2. 135 66=length of 7 sides=the perimeter. 

3. 20=apothem. 

4. 10=4 of 20=half the apothem. 

5. 1356.6=10Xl35.66=product of perimeter by half the 
apothem. 

.*. The area of the heptagon is 1356.6. 

Prob. XVII. To find the area of a regular polyg-on. when 
the side only is given. 

*Rllle. — Multiply the square of the side of the polygon by the 
nu7nber standing opposite to its name in the following table of 
areas of regular polygons whose side is 1 : 



II. 



III. 





Name. 


Sides. 


Multipliers. 


Triangle, 


... 


3 


^/3 = .4330127. 


Tetragon, or 


square, 


4 


1 = 1.0000000. 


Pentagon, 


. 


5 


|V1+|V5 = 1-7204774. 


Hexagon, 


- 


6 


h/B = 2.5980762. 


Heptagon, 


- 


/ 


1 cot. ifoo == 3.6339124. 


Octagon, 


- 


8 


2+2V2 = 4.8284271. 


Nonagon, 


- 


9 


£ cot. 20° = 6.1818242. 


Decagon, 


- 


10 


fV5-f2V5 = 7.6942088. 


Undecagon, 


- 


H i 


ycot. , 1 fi r °°= 9.3656399. 


Dodecagon, 


. 


12 


3(2+V3) =11.1961524. 



* Demonstration. — Since a regular polygon can be divided into as many 
equal isosceles triangles as it has sides, we may find the area of one trian- 
gle and multiply this area by the number of triangles, for the whole area. 
Let ABC be one of these isosceles triangles, taken from a polygon of ;/ 
s'des, AB and BC the equal sides, and A C the base. The angle at the ver- 
tex B— 360°-^-h. A=\(\m°— 360°-*-»)=C. From B let fall a perpendicu- 

BD 1 ^0° 

lar on A C at I). Then by trigonometrv, — — =tan (90°— — ). .\ BD= 

\AC n 

/1S0°\ 
\A C cot ( 1. The area of the triangle ABC~\A CXBD=\A C* cot 

/180°\ ' , ... . n /180°\ », L /180°\ 
l 1. /. The area of the polygonr=-yl C 2 cot I ) = ~ 9 cot I I 

where ^=side. By placing j=l, and »=13, 14, 15, &c, respectively, the area 
of polygons of 13, 14, 15, &c, side respectively, may be found. 



190 



FINKEL'S SOLUTION BOOK. 

I 



Prob. XVIII. To find the side of an inscribed square of a 
triangle, having- given the base and the altitude. 



Formula.— s-- 



ab 



the altitude. 



■+*' 



where ^=side, b the base, and a 



*Rul<3. — Divide the product of the base a?id altitude by their 
sum. 

I. What is the side of an inscribed square of a triangle 
whose base is 14 feet and altitude 8 feet? 



I* 

11.^3. 



T> C ""' 1 ab 14 >< 8 K-l' C 4.' ' 

By formula, ^__ = ___ = 5 T i T feet. 

8 feet=the altitude. 

14 feet=the base. 

112 sq. ft.=14x8=the product of the base and altitude. 
4. 22 feet=14 ft. +8 ft.==their sum. 
.5. 5 T y feet=112-r-22=the product divided b)- the sum 



III. .'. 5 T y ft.— the side of the inscribed square. 

VI. CIRCLE. 

Prob. XIX. To find the diameter of a circle, having 
given the height of an arc and a chord of half the arc. 

Formula. — D=k 2 -±-a, in which /§=chord of half the 
arc and a=height. 

fRule. — Divide the square of the chord of half the arc by the 
height of the chord. 



* Demonstration. — Let ABC be any triangle whose base is b and altitude 
a. Produce AC to H, making CH—BD. At IF erect the perpendicular 
HG and make HG=BD. Draw 
AG and at C, erect the perpen 
dicular FC, and draw FK. Then 
KE=FC—EN, and KN is the re- 
quired inscribed square. For, in 
the similar triangles AUG and 
A CF, we have A II : GH : : AC : 
FC, or a-\-b:a::b:FC. By inver- 
sion, and then bv Division, a:b 
::a—FC:FC, or BI-.FC. In the 
similar triangles ABC and KBE, 
AC:KE::BD:BI, or BD:AC:: 
BI.KE. Whence a:b::BI:KE. hlb - 12 - 

:.BI:KE:\B1\FC. .'. KE=FC and the figure KN has its sides equal 
and its angles right angles by construction. Hence, it is a square. Q.E.D. 

^Demonstration. — Let AB=k, the chord of half the arc ABC, and BD 
=a, the height of the arc ABC. Draw the diameter BE and draw the 




MENSURATION. 



191 



I. What is the diameter of a circle of which the height of 
an arc is 5 m. and the chord of half the arc 10 m.? 

By formula, D = k*~-a=10 2 -T-5= 

20 m. 

1. 10 m— AB, the length of chord 
of half the arc. 

2. 5 m.=BD, the height of arc. 

3. 100 m. 2 =square of chord. 

4. .-. 20 m.=100-f-5=BE, the diam- 
eter of the circle. 



II. 



III. .'. The diameter of the circle is 20 
meters. 




FIG. 13. 



Prob. XX. To find the height of an arc, having- given the 
chord of the arc and the radius of the circle. 



Formula.— a— R— si r< 

=.V the chord. 



in which, J?=radius and c 



*R,ule. — From the radius, subtract the square root of the dif- 
ference of the squares of the radius and half the chord. 

I. The chord of an arc is 12 feet and the radius of the circle 
is 10 feet. Find the height of the arc. 

By formula, a—R — y'/? 2 — c 2 =10 — y 10 2 — o 2 =2 ft. 

1. 10 ft.=the radius of the circle. 

2. 100 sq. ft.=square of the radius. 

3. 12 ft.=the chord. 

4. 6 ft.=half the chord. 

5. 36 sq. ft.=square of half the chord. 

6. 8 ft— y/]00 — 30 = square root of the difference of the 
squares of the radius and half the chord. 

7. .-. 10 ft. — 8 ft.=2 ft.=height of the arc. 

.-. The height of the chord is 2 feet. 



IT. 



III. 



Prob. XXI. To find the chord of half the arc, having 
$ven the chord and height of an are. 



Formula, — k=jja* + 



radius AO. The triangles ADB and BAE are similar, because their an 
gles are equal. Hence, BE\AB\\AB\BD, or BE.kwk.a. Whence BE 
— D—k-'i-i-a. Q.E D. 

N. B.— (1) If « and D are given, &=*j£>Xa\ (2) if D and k are given 

* Demonstra tion. — In Fig. 13, we have BD=BO—DO. But Z?Oz= 
SJA (T—DA ^-.-s^/r—c*]. .■. a=Ji— t/B*—c*. If a and R are given, (1) 
2c=2VI^ 2> -<' ff --«) 2 ] = 2V(2aiP— a 2 ); if a and 2c are given, (_>) B= 
(«*-|-c 2 )-*-2«. 



IWj 



192 FINKEL'S SOLUTION BOOK. 

*Rllle. — Take the square root of the sum of the squares of the 
height of arc and half the chord. 

I. Given the chord=48 ; the height=10, find the chord of 
half the arc. 

By formula, k=^ a 2 -\-c 2 =Vl0 2 +24 2 =V676==26. 

1. 48=the chord. 

2. 576=1 of 48 2 =square of half the chord. 

3. 10=height of chord. 

4. 100^= < quare of height of chord. 

5. 676— 576+100=sum of square of half of chord and 
height. 

6. 26=V676=square root of sum of square of half of chord 
and height. 

III. ' .-. The chord of half the arc is 26. 

Prob. XXII. To find the chord of half an arc, having- 
given the chord of the arc and the radius of the circle. 

Formula. — k=V%R 2 —jR\/4R 2 —4c 2 . 

j-Rule. — Multi-ply the radius by the square root of the differ- 
ence of the squares of twice the radius a?id the chord; subtract 
this product from twice the square of the radius and extract the 
square root of the difference. 

I. Given the chord of an arc=6 and the radius of the circle 
=5, find the chord of half the arc. 

By formula ± ^=V / 2i? 2 — i?\/4/? 2 — 4c 2 = / V / 2x5 2 — 5>/4X5 2 — 6* 
=VlO. 

1. 5=the radius of the circle. 

2. 10— twice the radius of the circle. 

3. 100=square of twice the radius. 

4. 6=chord of the arc. 

5. 36— square of the chord. 

6. 100 — 36=64=difTerence of squares of twice the radius 
and the chord. 

7. 8= / v / G4=square root of the above difference. 

8. 40=5x8=the product of the above square root and the 
radius. 

9. 50=2x5 2 =twice the square of the radius. 

10. VoO— 40=VlO= c hord of half the arc. 
III. .*. The chord of half the arc is V'To. 

* Dem onstration. —In Fig. 13, we have AB= yK /(AD 2 -\-BD 2 )=*J~[c 2 -\-a 2 ]= 
V / « 2 +c 2 . :.k=s]a 2 -\-c 2 . If h and 2c at e given, (l)a=^l- 2 —c 2 ; if k and a are 
given, (2) 2c=2\/k 2 —a 2 . 

^Demonstration. — From Prob. XXI., we have 7t=\ l a 2 -\-c 2 . From Prob. 
XX. we have a=R—\]R 2 —c 2 . ;. a 2 — 2R 2 — c 2 — 2R\J R 2 — c 2 . Sub sti- 
tuting this value of a 2 in the above equation, £=V 2x?" — ^v\4:R 2 — 4c 2 . 



II. 



MENSURATION. 193 

Prob. XXIII. To find the side of a circumscribed polygon, 
having- given the radius of the circle and a side of a simi- 
lar inscribed polygon. 

9 FCR 

Formula, — K '== . - , in which K is the side of 

V47? 2 — K 2 
the circumscribed polygon and K the side of a similar 
inscribed polygon. 
Rule. — Divide twice the product of the side of the inscribed 
polygon and radius by the square root of the difference of the 
squares of twice the radius and, the side of the inscribed polygon. 
I. When Ji=l, find one side of a regular circumscribed do- 
decagon. 

2KR 2K 

By formula, K = , . The formula does 

V4i? 2 — k* \k—K 2 
not lead to a direct result, since K is not given. But 
by the formula of Prob. XXL, if k is replaced by K 
we have A~=V / 2 — V-4 — 1 lor 2c=l, since it is the side 
of a regular inscribed hexagon, and K=V2 — ^3, since 
2c is a side of a regular inscribed dodecagon. 

2K 2V2— V3 V2— V3 

VII. RECTIFICATION OF PLANE CURVES AND 
QUADRATURES OF PLANE SURFACES. 

1. To Rectify a Curve is to find its length. The term 
arises from the conception that a right line is to be found which 
has the same length 

2. The Quadrature of a surface is finding its area. The 
term arises from the conception that we find a square whose 
area is equal to the area of the required surface. 

The formula for the rectification of plane curves is 

s =J**Jdx 2 -\-dy 2 =J >J j 1-f-I^J f dx, when the curve is re- 
ferred to rectangular co-ordinates. 

*=/4lT+»- [£) ' J *. or *=/J j r»+(*) ' J ^, when the 
curve is referred to polar co-ordinates. 



-/4 



■+(; 



■/JM 



are formulae for the rectifica- 
_1 2 _j_f^fl 2 ? c / v I tion of curves of double cur- 
ly) Ydy) S | vature, when referred to rec- 
tangular co-ordinates). 



a ■+(£)> 



194 FINKEL'S SOLUTION BOOK 

d6 
dr 






are formulae for the 
rectification of 
curves of double 
curvature, referred 
to polar co-ordi- 
nates. 



A= Cydx or Cxdy is the formula for the quadrature of any 
plane surface referred to rectangular co-ordinates. 

A^=: C\r 2 dO is the formula for the quadrature of plane surfaces, 
referred to polar co-ordinates. 

3. A Surface Of Revolution is the surface generated 
by a line (right or curved) revolving around a fixed right line as 
an axis, so that sections of the volume generated, made by a plane 
perpendicular to the axis are circles. 



$==27tjy~\ 



( dy^ 2 
1-|- | ~^= \ \ dx is the formula for a surface of revo- 



^dx 



lution, referred to rectangular co-ordinates. 

S=2tt fyds=27t ir sin $\ r 2 + j -U " \dd is the formula for 
a surface of revolution, referred to polar co-ordinates. 

V-=n Cy 2 d.x or x 2 dy is the formula for the volume of a solid 
of revolution referred to rectangular co-ordinates. 

V= f f Cdxdydz and V= C Czdxdy are formulas for the 

cubature of solids, requiring triple and double integration. 
// 
V= f CzrdOdr and V= C C Cr 2 sin ddcpdddr are the formulae 
for cubature of solids referred to polar co-ordinates. From 
the equation to the surface of the solid, z must be expressed 
as a function of r and 6. 

x 2 -\-y Q =I? 2 is the rectangular equation of a circle referred to 
the center. 

y 2 ^=2Rx — x 2 is the rectangular equation of a circle referred to 
the left hand vertex as origin of co-ordinates. 

r= k 2R cos. 6 is the equation of the circle referred to polar co-ordi- 
nates. 

Prob. XXIV. To find the circumference of a circle, the 
radius being given. 



MENSURATION. 195 

JFormuta.-C.=*f \J j 1+gj ' J dx=if B 
f x 2 +y 2y/ ~n —Rdx 1 1.3 1.3.5 , 

i^^-+&c)==4i?Xl.570796+— 3.141592X2^=2^^, in which 
,z.4.o.o.y 

7r=3.141592-f-. Since the diameter is twice the radius, we have 

27tR=7z£>, in which D is the diameter. .'. C=27tR=7tD. 

C C 

.'. (1) R=rzr^—, (2)D= — , where C is the circumference. 

Rule. — Multiply twice the radius or the diameter by 3.14-1592. 

I. What is the circumference of a circle whose radius is 17 
rods? 



11.(2'. 



By formula, C=27r7?=3.141592 X 34 rods = 106.814128 
rods. 

1. 17 rods=the radius. 

34 rods=2xl7 rods=the diameter. 

106.814128 rods=3.141592X34rods=the circumference. 



III. .-. The circumference is 106.814128 rods. 

Note. — The ratio of the circumference to the diameter can not be exactly 
ascertained. An untold amount of mental energy has been expended upon 
this problem; but all attempts to find an exact ratio have ended in utter 
failure. Many persons not noted along any other line, claimed to have 
found this clavem imfossibilitibus by which they have unlocked *11 the diffi- 
culties that have encumbered the quadrature of the circle for more than two 
thousand years. The Quadrature of the Circle is to find a square whose area 
shall be exactly equal to that of the circle. This can not be done, since the 
ratio of the circumference to the diameter can not be exactly ascertained. 
Persons claiming to have held communion with the "gods" and extorted 
from them the exact ratio are ranked by mathematicians in the same class 
with the inventors of Perpetual Motion and the discoverers of the Elixir of 
Life, Alkahest, the Fountain of Perpetual Youth, and the Philosopher's 
Stone. Lambert, an Alsacian mathematician, proved, in 1761, that this ratio 
is incommensurable. In 1881, Lindemann, a German mathematician, dem- 
onstrated that this ratio is transcendental, and that the quadrature of the 
circle by means of the ruler and compass r only, or by means of any algebraic 
curve, is impossible. Its value has been computed to several hundred deci- 
mal places. Archimedes, in 287 B. C, found it to be between 3)? and 3}; 
Metius, in 1640, gave a nearer approximation in the fraction \\\ ; and, in 1873, 
Mr. W. Shank presented to the Royal Society of London a computation ex- 
tending the decimal to 707 places. The following is its value to 600 deci- 
mal places: 

3. 141,592,653,589,793,238,462,643,383,279,502,884, 197, 169, 
399,375,105,820,974,944,592,307,816,406,286,208,998,628,034,825, 
342,117,067,982,148,086,513,282,306,647,093,844,609,550,582,231, 
725,359,408,128,481,117,450,284,102,701,938,521,105,559,644,622, 
948,954,930,381 ,964,428,810,975,665,933,446,128,475,648,233,786, 



196 FINKEL'S SOLUTION BOOK. 

783,165,271,201,909,145,648,566,923,460,348,610,454,326,648,213. 
393.607,260,249,141,273,724,587,006,606,315,588,174,881-520,920, 
962,829,254,091,715,364,367,892,590,360,011,330 530,548,820,466. 
521,384,146,951,941,511,609,433,057,270,365,759,591,953,092,186, 
117,381,932,611,793,105,118,548,074,462,379.834,749.567,351,885, 
752,724,891,227,938,183,011.949.129.833.673.362.441,936.643086, 
021,395.016,092,448,077.230,943,628.553.096,6:20,275,569,397,986, 
950,222,474,996,206,074,970,304123,669+. 

i 7r 2 =— -,-)-— -—]-— -(-—-f-^c* Bernoulli's Formula. 
1 " "Z" o 4 

^tt= - r ~i_-^-^- p— .... vv Mllis s Formula, 1655. 

L3 5 7-9. 11-. ..&c. J 

|7r=l-|-l Sylvester's Formula, 1869. 

1+^3 
l-f3\4 

1+4.5 

1+ 
4 
— =1+P_ 

* 2+3^ 

2+5^ 
2+7^ 

2-f- Buckner's Formula. 

The Greek letter n, was first used by Euler, to designate the 
ratio of the circumference to the diameter. 

Prol). XXV. To find the length of any arc of a circle, 
having given the chord of the arc and the height of the arc, 
i. e., the versed sine of half the arc. 

(a). Fornmla.-s=f4l+ gj ' dx== J\^±p^ dx = 

f Rdx . x D \-x x » 1.3** , 

1.3.5« 7 . „ "I 1.,, ..rV»— a"' , ,(<: 2 — a«-y 
2.4.6.7.^. + &C J '=2~J a '+ c > bq^+%^RP + 

& (a*+*2 )5 +rf 2(^2^2 J +&c^, in which «=the altitude of 
the arc and c=half the chord of the arc. 

*Note. — This series was discovered by Bernoulli, but he acknowledged 
his inability to sum it. Euler found the result to be^7r 2 . For an inter- 
esting discussion of the various formulae for tt, see Squaring the Circle^ 
Britannica Encyclopedia 



MENSURATION. 197 

(I. ) Formula— s=arc=2^( a 2 +r 2 ) X l+ ,./> „ * 

This formula is a very close approximation to the true length 
of the arc when a and c are small. The first formula may be ex- 
tended to any desired degree of accuracy. 

Rule from (b). — Divide 10 ti?nes the square of the height of 
the arc by 15 times the square of the chord and 33 times the height 
of the chord; multiply this quotient increased by 1, by 2 times the 
square root of the su?n of the squares of the height and half the 
chord. 

I. The chord of an arc is 25, and versed-sine 15, required the 
length of the arc. 

vt r i "/ \ a*+c* r c2—a* , x (c*—a* y , 3 

By formula (a), «rc=^— \^^H (^^ J + 4 J } 

( C 2_ a 2 y 5 ( C 2_ a y y -j_ 15 2 +25 2 T 25 2 - 15 2 , 1 

!* 2 +c 2 J +112 [« 2 +c- J + &C J — 2X15 Ll5 2 +25 2+ (J X 
f 25 2 -15 2 ]3 , 3 f 25 2 -15 2 V , 5 f 25»—15» V , Sr -i w _ 

ll^+^J +i:ilP+2Pj +n2ll5M3^J + &c J= 53 - 68 

+ft. 

1. 25 ft.=length of the chord. 

2. 15 ft.=height of the arc, or the versed-sine. 

3. 2250 sq. ft.=10 times 15 2 =10 times the square of the 
height of the arc-. [chord. 

4. 9375 sq. ft. =15 times 25 2 =15 times the square of the 

5. 7425 sq. ft. =33 times 15 2 =33 times the square of the 
height of arc. 

II <j 6. 17*100 sq. ft.=7425 sq. ft.+9375 sq. ft. 

7. ^=2250-s-17800=10 times 15 2 -f-(15 times 25 a +33 

times 15 8 ). 
8- l-K 4 A=ttt= 1 + 1 times 15 2 -^(15 times 25 2 -f33 

times 15 2 ). 
9. 3811 sq. ft.=15 2 +(12j) s . 

10. 53.5S ft. = f^xVl5 2 +(12i) 2 =MiX|V61=length of 
arc, nearly. 

III. .-. 53.58 ft.=length of the arc. 

Prob. XXYI. To find the area of a circle having given the 
radius, diameter, or circumference. 



Formula.— A=A f ydx,=\ f '(7? 2 — x ~)y i dx=\Vj I x 

x — I R l~X X S 1 2 V 5 

^T^i+ &c =i 7r J l =*K*=\*D-= < ^=yi-X. C, when the ra 



198 FINKEL'S SOLUTION BOOK. 

dius and circumference are given. .*. (1) J?=\A-t-?T, (2) D= 
V4^h-tt=2 R=2V.4-j-tt, and (3) C=ViS=2V^Z. 

Rule I. — 7%^ #r£# <?/* « circle equals the square of the radius 
multiplied by 3.1J/.1592; or (2) the square of the diameter multi- 
plied by .785398; or (3) the square of the circumference multiplied 
by .07958/ or (J/.) the circumfere?ice multiplied by \ of the diame- 
ter; or (5) the circumference multiplied by -J of the radius. 

Rule II. — Having given the area. (1) To find the radius: 
Divide the area by 3.1J/.1592, and extract the square root of the 
quotient. (2) To find the diameter: Divide trie area by 3.1-^1592 
and multiply the square root of the quotient by 2. (3) To find 
the circumference: Multiply the area by 3.1J/.1592 and multiply 
the square root of the product by 2. 

I. What is the area of a circle whose radius is 7 feet? 

By formula, .4=7r J ff 2 =3.141592x7 2 =153.93804 + sq. ft. 

A 7 ft.=the radius. 
IIJ2. 49 sq. ft.=7 2 =square of the radius. 

IS. 153.93804 sq. ft.=3.141592 X49 sq. ft.=area of the circle. 

III. .'. 153.93804 sq. ft.=area of the circle. 

I. What is the area of a circle whose diameter is 4 rods? 

By formula, ^4=iMrZ> 8 =ix3.141592x4 8 =12.566368 sq.ft. 

rl. 4 ft.=the diameter, 
-r-r J 2. 16 sq. ft.=square of the diameter. 

|3. 12.566368 sq. ft.=±X 3.141592 X4 2 =.785398x 16 sq. ft. 
t =area of the circle. 
III. .*. 12.566368 sq. ft.=area of the circle. 

I. What is the area of a circle whose circumference is 5 meters? 

C 2 25 2 
By formula, ^=— =-f-==1.989 m 2 . 

{1. 5 m.— the circumference. 
2. 25 m. 2 =the square of the circumference. 
3. 1.989 m. 2 =.07958x25 m. 2 =the area of the circle. 
III. .'. 1.989 m. 2 =the area of the circle. 

Remark. — We might have found the radius by formula (1) under Prob. 
XXIV and then applied the first of Rule I. above. We might have found the 
radius by formula (1) of Prob. XXIV and then applied (5) of Rule I. above. 

I. What is the circumference of a circle whose area is 10 A. ? 



MENSURATION. 199 



By formula (3), C=2\^4=2V3.141592 X 1 600=80^60 X 
1.7724539=141.796312 rods. 

1. 10 A. =1600 sq. rds.=the area of the circle. 

2. 1 600-7- 7T=the square of the radius. 

3. ^ 40 JLj^r=the radius. 



II, 



a 80 /- 40 r- 

4- — V^r=7r2 times ~VJT=the diameter. 

7T 7t 

40 - 
5. 80V7T=7rX ~V7T=141. 796312 rocls=the circumference. 

III. .'. 141.796312 rods=the circumference of the arcle. 

I. With what length of rope must a horse be trM to a stake 
so that he can graze over one acre of grass and no more? 

By formula (1), 7?=^-T-7r=V160-j-7r=4 > ]— =7.1364+ rd. 



II. 



1. 1 A. =160 sq. rd.=area of the circle over which the 
horse can graze. 

2. 160-j-7T=square of the radius. 

3. Vl 60-7- 7r=4\ / 10-^7r=7. 1364 rd.=radius or length of rope. 

III. .*. 7-1364 rd.=lcngth of the rope. 

Prob. XXVII. To find the area of a Sector, or that part of 
a circle which is bounded by any two radii and their included 
arc, having" given the chord of the arc and the height of the 
arc. 

*Forniula.—A= jydx=2 \ (R 2 —xS)V*dx=x(R 2 —x 2 )x 

+i? sin ^=-^r\{-^r\ ~{-*r\ \ + {-£r 
^~ l ^+^= c [-^r\ + r^r] sin ^+^T C (n>H 

-r-[ 2a J { <:»+«»" t "l-2.3(/:»+««J ^1.2.3.4.5 U 2 -H 2 j " r 

1 9 34 5 6^ U 2 4-g g ' ~^~ &C ^ in which c is half the chord of the 
arc and a the height of arc. 

Demonstration. — Let A />'— \\ BD=.y, and It=AD=^th& Tadius of-the oir- 
cle. Then x 2 -\-y 2 =ft 2 , the equation of the circle referred to the center. 
Now A— 2 Cydx; but v=(7? 2 — .\" 2 )/^, from the equation of the circle. 

.-. A=2 f(# 2 — * 2 )Xdx=x (A> 2 — .v 2 )X+7? 2 si» i j,- But x=7?— a and y= c. 
Hence A=(X— rt)[A> 2 — (A— ,,)2]-f-7?2 sin '-"". But, from (2) Prob. XX, /? 



200 



FINKEL'S SOLUTION BOOK. 



Rule. — (1) Find the length of the arc by Problem XXV, 
and then multiply the arc by half the radius which may be found 
by Problem XX, in which c and a are known and R is the un- 
known quantity. 

(2) If the arc is given in degrees, take such a part of the 
whole area of the circle as the number of degrees in the arc is 
•fS60». 

I. Find the area of the sector, the chord of whose arc is 40 
feet, and the versed-sine of half the arc 15 feet. 

By formula, A=c (^=Sl) + f"!±f!] 2 \ ^^1 + J^ 
y { 2a )^{ 2a J ) c*+a*^l .2 .3 



( c 1 — <*y '" r 20 2 — 15M . rl5 2 +20V c 20 s 



—15' 



1 f20 2 — 15 2 U 



— ( 

1.2.3 { 



IU 



+15 2 ' 

T20 2 -15 2 ^ 5 

20 2 +15 2 J ' 1.2.3A5(20 2 +15 2 J + &c -= 558 - 12 5 s q- ft - 

1. 53.58 ft . = 2(15^20^)Hx;i+ 6()x2 1 ^g xl5 ^ 

length of the arc, by (b), Prob. XXV. 

2. 20f ft.= — ^- =radius of the circle, by solving the 



formula of Prob. XX with respect to R. 
3. .'. 558.125 sq. ft.=£ (20|-X5'3.58)=area of the sector. 

III. .'. 558.125*sq. ft.=the area of the sector. 



a 2 +c 



+C 2 __ . /c-2— «2\ /c 2 — « 2 \ 2 . / C 2_«2\ 



Trigonometry, we have sirr^fcrrfl— -— 6 3 | fls_& c Hence, ^4 =z 

r c Z_ g 2 -> r C ?_ q» -[ 2 f C 2— rt 2 J^ j r 2_ ^ \ 3 , __1___ fc^ff 1 5 „ ) 

[ 2a J C+ V 2a J < c 2 +a* 1.2.3 U*+« 2 J i "l.2.3.4\5 Ic 2 -^ J ~ &c -$ 



In this formula 



•('-¥)" 



is the area of the tri 



_a a +c 3 



angle DEA. For *=./? — az_ 

& 2a 2a 

the altitude and c is half the base of the triangle 

re 2 a 2 } 

.-. j — j c — the area of the triangle DEA. 

Therefore, if we subtract the area of the tri- 
angle DEA from the area of the sector,, we 
shall have the area of the segment DEC. Hence, 

the area of the segment DEC is j — - — I 

{(3$) -is (Sr+ojbs (S3-M • This result ™y 

be carried to any desired degree of accuracy. 




FIG. 74. 



MENSURATION. 201 

I. What is the area of a sector whose arc is40°and the radius 
of the circle 9 feet? 

(1. 9 ft.=radius of the circle. 
2. 7ri? 2 =7r9 2 =area of the whole circle. 
IU3. 40°=length of the arc. 

4. 40°=4 of 360°. 

5. tt9=4 of tt9 2 =28.274328 sq. ft.=area of the sector. 
III. .-. The area of the sector is 28.274328 sq. ft. 

Prob. XXVIII. To find the area of the segment of a circle, 
having- given the chord of the arc and the height of the 
segment, i. e., the versed-sine of half the arc. 

xr t t \ a f c2 — a2 V \ {c 2 —a*\ 1 fc 2 — a*y 

Forniula.-(a)A=[-^-\ J ^__J__J__j 

*Rllle. — Divide the cube of the height by twice the base and in- 
crease the quotient by two-thirds of the produt t of the height and 
base. 

I. What is the area of a segment whose base is 2 feet and al- 
titude 1 foot? 

By formula (*), ^-^ + |-(9^) = ^ + |(2 X l)=l T Vsq.ft. 

* Demonstration - In the last figure, let 2?C=tf=altitude of thesegment and 
DE=2c— the base of the segment. Then BD*=BCxBF=a (2/?— «)=c 2 . 

Whence R=-4^- . and AD—R—a— 



r 2 —a 2 /— — 

a= — ■ — . DC=k—V c 2 -\-a 2 . 
% 'n 



2a ' 2a 2a 

T> 7~) r 

By Trigonometry,—— =sin / DAC, or-= sin/ DAC. Now 2tt/?:=3(>0 o . 

_, 180° _. , __ __ 180° , circDC 180° T . 

.-./?=: Therefore, i?: arcDC:: : ?= - — X Let s = arc 

TT TT R TT 

DCE. Then the /_DAC— / ^—=zl 7 —-. .". — — sin— -. In like manner, from 

DC 

the right angled triangle T^C, — -=sin / CFD, or since the / C^Z>=the 

k s 1 

% /_ CAD, 7r^=sin- : — . Now since the sine of any angle 6=6 6 s 4- 

2R AR 1.2 3 

■ r # s — ^ 7 -|-&c., the above equation becomes 

1.2.D.O 1.2.Q.O 7 

c s 1 ( s \ z 1 f * 1 5 

»"53r-i«lf»J +r^35l.2^J - &c (1) - and 

k s 1 { i \ 1 f O 5 

2^ = Tff~ir2^l4y?J 3+ L2T5l4^J ~ &C (2) ' Multiplying equation (2) 

by 8 and subtract equation (1) in order to eliminate the term containing 

s*, we have approximately,—^— r=^— % 1 ] ^ g . J I — I -f &c. Omitting 

the negative quantity, since it is very small in comparison with .s and be- 
cause it is still more diminished by a succeeding positive quantity, we have 



II. 



202 FINKEL'S SOLUTION BOOK. 

-1. 1 ft.=altitude of the segment. 

2. 2 ft.=base of the segment. 

3. 4 ft.=twice the base of the segment. 

4. 1 cu. ft.=cube of the height of the segment. 

5. \ sq. ft.=l-^4=quotient of the cube of the height and 
twice the base. 

6. 2 sq. ft.=2Xl=P ro d uct of the height and base. 

7. H sq. ft.=f of the product of the height and base. 
-8. .'. 1-^- sq. ft.-|-^ sq. ft.=l T 7 ^sq. ft.=area of the segment. 

III. .*. The area of the segment is 1 T 7 ^- sq. ft. 

Prob. XXIX. To find the area of a circular zone, or the 
space included between any two parallel chords and their 
intercepted arcs. 



1.2.3.4.5 



-^ T , . . \c*—a*Y c c*—a* 1 [c^—an 

Formula.-{a) A= j-^— j j ^+a~-T^ \7^\ 

— a 2 ] 5 ) _ fc 2 —a^] 2 ( c^—d* 1_ 



J ^1.2.3.4.5 U /2 +« /2 j 



-&c. ( 



\c'*-\-a' j i.zj.o.i.tj \y -pc* - 

Rule. — Find the area of each segment by Prob. XXVIII., and 
take the difference between them, if both chords arc on the same 
side of the center; if on opposite sides of the center, subtract the 
sum of the areas of the segments from the whole area of the circle. 

I. What is the area of a zone, one side of which is 96, and 
the other 60, and the distance between them 25? 

Let AB=60=2c / , CZ>=9Q=2c, and HJ^=25=k. Then AH 
=30=*:' and CK=iS=c . Let O A=,R. Then 
OH=sj_R2_ c ^ i QX =sJR^—c -. But OH 
= OX+ HK\ .-. Vi?2_^ 2= Vi?2_ c a _|_ h. 

Whence R=^r *l±c*h* + (c*—h*—c'*y. 
.-. LH=a / =R—OH=R—sJ(R' 1 —c^)=-} 

lj±C*—h* -f { C 2 =k 2 — C' 2 ) 2_lV4/ 2 2 ( C 2 

g£ 2c S^rM^T 2 2« / 

»S= — — — — — =^'(4v c 2 -\-a 2 — a). This is the approximate 

o o 

length of an arc in terms of its height and base. Now the area of the seg- 
ment DCE—% A CXarcDCE— area of the triangle DEA—% RXS—% AB 

-f 2c«2]z=i[ v /(16r 6 +48c 4 « 2 +48c 2 «4_^iGa6)_4c 3 +2c^]=-^ | (4a 3 -f-6c 2 «+ 
|^)„earl y -4c3 + 2,^]=^ (s ? +|p] =^+%(2 C «). «. E. D. 




MENSURATION. 203 

J r ( K ci_i l -2_ c "iy % i n like manner, LjC=a=zR—*lR* — c 2 ,=— - 

.-. By formula (3), ^=^+f <2«)-[ g ^j+K2^a')] ,= 
jl|V4 C ^«+( C 2 -A 2 -^ 2 -(^-AV")^j.V2.2 C +}x2 C X^ 

rV^^qr( C 2_^2_^2)2_( c 2_/ 2 2_ c /2)2-|_| ( i\^¥+p 
_^2-ZZ7r)2_V 4A -2( c 2__ c /8) + ( c 2_/,2_ c /2)?J j* _i_0(2c)+t X2/ 

xiy[V4/ 2 2 ^ + (6- 2 -^ 2 — 6- /2 ) 2 — V / 4^2( 6 -2_ 6 .^) + (,2_ / ,2_ < /2 ) 2^| 
2/2 / 

=2547— 4084=2138}. 

1. 50=i?=^V / 4 c V^-+(c 2 -^^^^=^V / 4 X 48 2 X 25* + 



II. 



(48 8 — 25 2 — 30 2 ) =radius o f the circle. 

2. OK=<JR 2 —c 2 =V50 2 — 48 2 =14. 

3. .'. Zjff"=a=50— 14 =36=alti tude of segment CZ£>. 

4. OH=^R '^=7 *~=V50 2 —80 2 =40. 

5. .. ZZ r =a / =50—40=10=altitude of the segment ylZ/;>\ 

9fi3 

6. .-. — f-f(96 X 36 )=2547=area.of segment CDBLA. 

10 :i 

7. - __|_$(60X10)=408£=area of the' segment ^Ii?Z. 

8. .-. 2547— 408^=2 l;>S' r ;=area of the zone CDBA. 



III. .-. 2138 2 r=area of the zone ABDC. 



Note. — This result is only approximately correct. The radius of the cir- 
cle may be found by the following rule: 

Subtract half the difference between the tzvo half ektrds from the greater 
half-chord^ multiply the remainder by said difference, divide the product by 
the width of the zone, and add the quotient to half the width. To the square 
of this sum add the square of the less half chord, and take the square root of 
the sum. 

This rule is derived from the formula in the above solution, in which 
7?= 1 ^+p-^-cT) 2 ^ jr (r^+// 2 -f^) 8 +4// 2 c^ l 



204 



FtNKEL'S SOLUTION BOOK. 



Prob. XXX. To find the area of a circular ring-, or the 
space included between the circumference of two concen- 
tric circles. 

FormulSL* — (a.) A=n (B 2 — r 2 ), in which B and r are 
the radii of the circles. 

(b.) *A=%7ic 2 y in which c is a chord of the larger circle 
tangent to the smaller circle. 

I. Required the area of a ring the radii of whose bounding 
circles are 9 and 7 respectively. 

By formula (a), A=:7r(B 2 — r") = 7t(9 2 —7 2 )=S27r= 
100.530944. 
9=i?=radius of the larger circle, and 
7=r=radius of the smaller circle. 
7r9 2 =7T^t >2 =area of larger circle, and 
7r r J 2 =7rr 2 =arca ot smaller circle. 

... 7r 92__ 7r7 2 =: .-, 7r (U I 2_72) == 327r= 

100.530944=area of the ring. 
. 100.530944=the area of the ring. 



II. 



III. 



* Demonstration. — Let ABC be the chord of the 
large circle, which is tangent to the smaller 
circle, and let ABC—c. Then BC=%a= 



J(OC 2 — 0^2)z= v (i?2— r 2 ). 



\c*=R'< 




}ttc 2 — vr(i?2— r 2 ) But7r(i?2— r 2 ) is the dif- 
ference of the areas of the two circles or the area 
of the ring ". \irc 2 z=zthe area of the ring. 
Q. E. D. FIG - 16 > 

Prob. XXXI. To find the areas of circular lunes, or the 
spaces between the intersecting arcs of two eccentric circles. 

a'* r- n'* 

Formula. — A= „, n x - |-| (2ca)- 



segments of which the lunes 



2(2c) 

Rule. — Find the area of the two 
are formed, and their difference will be the area required. 

I. The chord AB is 20, and the height DC is 10, and BE 2; 
find the area of the lune ARBC. 



-JCfTS^H^Hlt 



C — C 



(c-S) 



-#] 



-K : 



If now we find the altitudes of the two segments and then find the length 
of the arcs of the segments by formula (b), Prob. XXV, and then find the 
area of the sectois by multiplying the length of the arcs by half the radius, 
from the areas of the sectors subtract the triangles formed by the radii of 
the circles and the chord of the arcs, we shall thenhave the area of the two 
segments. Taking their difference, we shall have for the area of the zone 
2136-75, which is a nearer approximation to the true area. 



MENSURATION. 



205 



By formula, ^--^ +t (2^)-[^y-H(2^)]=- 
+|(20Xl0)-[^ Q +f(20X2)]=131 T V 



IU 



III. 



1. .^=chord=20. 

2. BE = height of segment 

AEBD=% [ACBD=10 

3. DC = height of segment 

4 -2So + * (20Xl0! ^ 158i = 
area of the segment A CBD. 

5. J^U-r-f (20X2) =26if= 




F7G. 17. 



2X20 
area of the segment AEBD. 

1584 — 26|!=131 T V=area of the lune ACBE. 



VIII. CONIC SECTIONS. 



DEFINITIONS. 



1. The Conic Sections are such plane figures as are 
formed by the cutting of a cone. 

2. If a cone be cut through the vertex, by a plane which also 
cuts the base, the sections will be a tyianglc. 

3. If a right cone be cut in two parts, by a plane parallel to 
the base, the section will be a circle. 

4. If a cone be cut by a plane which passes through its two 
slant sides in an oblique direction, the section will be an ellipse. 

5. The Transverse Axis of an ellipse is its longest 
diameter. 

6. Hie Conjugate Axis of an ellipse is its shortest 
diameter. 

7. An Ordinate is a right line drawn from any point of 
the curve, perpendicular to either of the diameters. 

8. An Abscissa is that part of the diameter which is con- 
tained between the vertex and the ordinate. 

9. A Parabola is a section formed by passing a plane 
through a cone parallel to either of its slant sides. 

10. The Axis of a parabola is a right line drawn from the 
vertex, so as to divide the figure into two equal parts. 

11. The Ordinate is a right line drawn from any point 
in the curve perpendicular to the axis. 



206 FINKEL'S SOLUTION BOOK. 

13. The Abscissa is that part of the axis which Is con- 
tained between the vertex and the ordinate. 

13. An Hyperbola is a section formed by passing a 
plane through a cone in a direction to make an angle at the base 
greater than that made by the slant height. It will thus pass 
through the symmetrical opposite cone. 

14. The Transverse Diameter of an hyperbola, is that 
part of the axis intercepted between the two opposite cones. 

15. The Conjugate Diameter is a line drawn through 
the center perpendicular to the transverse diameter 

16. An Ordinate is a line drawn from any point in the 
curve perpendicular to the axis. 

IT. The Abscissa is the part of the axis intercepted be- 
tween that ordinate and the vertex. 

1. ELLIPSE. 

a 2 y 2 -\-b 2 x 2 -=-a 2 b 2 is the equation of an ellipse referred to the 

center. 

b 2 
y 2 =—(2ax — x 2 ) is the equation of the ellipse referred to left 

hand vertex. 

In these equations, a is the semi-transverse diameter and b the 
semi-conjugate diameter ;jy"is any ordinate and x is the corres- 
ponding abscissa. When any three of these quantities are given 
the fourth may be found by solving either of the above equations 
with reference to the required quantity. 

a (\ e 2 \ 

p=—-± '.. is the polar equation referred to the centre, and 

1 — e cos 6 

a ( 1 e 2 \ 

p= — ^ — t, is the polar equation referred to the left hand 

1 i~4-£ cos l H 

vertex. 

Prob. XXXII. To Unci the circumference of an ellipse, the 
transverse and conjugate diameters being known. 

Formula.— cir. = C=4f*/dy 2 +dx 2 =4: / >|^r( 1—c 2 ) 2 

. /V r 2 i^n— e 2 ) 2 C \a 2 —e 2 x 2 

dx 2 -\-dx 2 =4 J ^ { i dx=4 / J-^5 %-dx= 

J y •/ ^ a i — x £ 



4/ /-o o -N 1 r -dx=4\a sin -- -^-sin x 

Vo V a — x a v a 2a [2 a 

x i e4: f3(7 2 {a 2 . x x /— x 3 



MENSURATION. 207 



SUo" L 4 V 2 /J 2.46« 5 | 6 L 4 V2*T J y ~~ ; * / r 



r e 2 3^ 3.3.5.* 6 "1 . , . . i/a*—b 



2 ™i^i^f£u- 2lAAs.<r &c j in which te ^i— 

Ullle. — Multiply the square root of Jialf the sum of the squares 
of the two diameters by 3.1J/.1592, and the product ivill be the cir- 
cumference, nearly. 

I. What is the circumference of an ellipse whose axes are 24 
and t3 feet respectively ? 



By formula, C7r.= C=27rXl2 



2.2 V 12 2 >/ 



2 



II 



L 2T4V 1_ I2O "~ &C * = 27r Xl2X.87947.==66.31056 ft., nearly. 

1. 576 sq. ft.=24 2 =square of the transverse diameter. 

2. 324 sq. ft.=18 2 ==square of the conjugate diameter. 

3. 900 sq. ft.=sum of the squares of the diameters. 

4. 450 sq. ft.=half the sum of the squares of the diameters 

5. 1 5^2 ft.==V450=square root of half the sum of the squares 
of the diameters. 

6. 7T15V2 ft.= 66.6434 ft., nearly, =the circumference of 
the ellipse. 

III. .'. The circumference of the ellipse is 66.6434 ft. nearly, by 
the rule. 

Prob. XXXI 1 1 . To find the length of any arc of an ellipse, 
buying- given the ordinate, abscissa, and either of the diam- 
eters. 

r o 

&c (. — asm — -sin s l a i— x *\ — 

5 \ J a 2a{2 a T J 2.4« ;; L 4 

' flZ x X A "' — I 1 

— sin -1 - — ~\fd- — x 2 — j-V« 2 — x' 2 — &c, in which .v is the ab- 

La b% 

scissa; a the semi-transverse diameter ; and e=+A =the ec- 

N a' 1 

ccntricity of the ellipse. 



208 



FINKEL'S SOLUTION BOOK. 



Rule# — Find the length of the quadrant CB by Prob. XXXI 
and CF by substituting the value of x in the above series. Twice 
the ditference between these arcs will give the length of the arc 
FBG. 

I. What is the length of the arc FBG, if 0£=x=:9,FF=y=$, 
and <9C=£=10? 

Since a 2 y 2 -\-b 2 x 2 =a 2 b 2 , we find, by substituting the values of 
x,y, and b, a=15. Then by the formula, FBG=s=2 I \na \ 1 — 



(i) 



1 (2 4^ 

- .* c" 

sin — 

a la 



*e x 



4 _ri 3 5) 

a 2 . -x x i — - 1 

^2 a 2 J 

* rSa*(a* . * *,— 






2.4a* 

3 



l4a*—x*Q—kc. 



:^l 6 |i-( 4 ).L 



-(if) 2 3~(H-f) 2 f~&c. J -2)15^-^ 



2.15 1 2 15 2 V15 9 J 2Xi5^L"T [ 

V15 2 -9 2 — |Vl5 2 — 9*1 — &c. | =7tl5 



FIG. 18. 
fl5 2 



T sln 15 ' 



9 

2 

f 185 _ n , * 4 T3.15 2 r 185 

[_^72)- 8 — [_ ( _ 7r _72-3 7 ]-&c \ =38.406-. 

Prob. XXXIV. To find the area of an ellipse, the trans- 
verse and conjugate diameters being' given. 

Formula.-A=4Jydx=4 *JJj a , _ x 2 )dx==7ra ^ in 

which a and b are the semi-transverse, and semi-conjugate diame- 
ters. 

Mule. — Multiply the product of the semi -diameters by 7t= 
3.141592, or multiply the product of the diameters by \7T=.785398. 

I. What is the area of an ellipse whose traverse diameter is 70 
feet and conjugate diameter 50 feet? 

By formula, ^4=7r^=7r35X 25=2748.893 sq. ft. 

A. 35 ft.=4of 70 ft.=length of the semi-transverse diameter. 
II.<2. 25 ft.=| of 50 ft.=length of the semi-conjugate diameter. 
13. .". 2748,893 sq. ft=7rx3oX25=the area ofthe ellipse. 
III. .-. The area of the ellipse is 2748.893 sq. ft. 

Note.— 7ra£ = -/7r«2 tt£2. .-. The area of an ellipse is a mean propor- 
tional between the circumscribed and inscribed circles. 



MENSURATION. 209 

Prob. XXXV. To find the area of an elliptic segment, hav- 
ing- given the base of the segment, its height, and either di- 
ameter of the ellipse, the base being parallel to either diam- 
eter. 

Formulae. — (a)A= I ydx, or / xdy=~ fx(a 2 — x 2 )%dXy=. 

&)•-»&]'-« Iffl"--" (*] "^ 

(b) -4=. /^rf«=-.r*(a'— x s )i+a? sin" 1 -~|. 

The former formula of (<z) gives the area of a segment whose base 
is parallel to the conjugate diameter and the latter the area of a 
segment whose base is parallel to the transverse diameter. 

Rule. — Find the area of the correspo?iding segment of the 
circle described upon the same axis to which the base of the seg- 
ment is perpendicular. Tlien this axis is to the other axis as the 
area of the circular seg?nent is to the area of the cllcptic seg?nent. 

2. PARABOLA. 

y 2 =2px is the rectangular equation of the parabola. 
P 



1 — cos 6 



is the polar equation of the parabola. 



In the rectangular equation, HG=y, the ordinate; AG-=x, the 
abscissa; AF=AF=^p. If any two of these are given the re- 
maining one may be found from the equation, p is a constant 
quantity. 

Prob. XXXVI. To find the length of any arc of a parabola 
cut oft' by a double ordinate. 

Formula.— s=2j\Zdy 2 +dx 2 =-~ f (p*-\-yt) l A dy= 
y-VpH^+p \og[y+Vf^^]+C=Y¥+T+ 
P log (Z±!^±2' 2 ] oT^f<J>*+y*)X dy=<l (y+i. $£- 



210 



FINKEL'S SOLUTION BOOK. 



Rule, — When the abscissa is less than half the ordinate: To 
the square of the ordinate add f of the square of the abscissa and 
twice the square root of the sum will be the length of the arc. 

I. What is the length of the arc KAH, if^Gis2and GH §t 
By formula, s=^v / /> a -f-y 2 + 

^io g [d^S]Jv^ +/ log 






f 



]• 






Since y 2 =2px, we have 

.•.5=f(3Vra)+9 log 



r- 6+3Vl 3H 2 VT 3+ 9 log *(2+Vl3), 



or 




F/G. 79. 



by series, s- 



=2O+i-i^-i-i'H^+i-fH-*-^-^c.]=12.7105 

length of the arc, nearly. 

1. 2=A G=the abscissa. 

2. 6= GJ7=the ordinate. 
I ) 3. 36=lhe square of the ordinate. 

4. 1 3 6 =|- of 2 2 =-| of the square of the abscissa. 

5. 2V( 1 3 6 +36)=12.858=the length of the arc, nearly. 
Ill .-. 12.858=length of the arc, nearly. 

Prob. XXXVII. To find the area of a parabola, the base 
and height being- given. 

Formula. — A =2 / ydx=2 I -y 2 dy=-^- ==|xr =f (x. 2y ) , 
i. e., the area of parabola HKA is •§■ of the circumscribed rectangle. 

Rule. — Multiply the base by the height and § of the product 
will be the area. 

I. What is the area of a parabola whose double ordinate is 24m. 
and altitude 16m. ? 

By formula, A=$ (x. 2y)=}( 16X24 )=256mS. 

r l. 24m.=i7A r '(in last figure )=the double ordinate, or base 
of the parabola. 
16m. 



9 



, . r jz. iom.=i4 G— the altitude of the parabola. 

j3. .\384m 2 =:16x24=the area of the rectangle circumscribed 
about the parabola. 

[4. § of 384m 2 =256m 2 =the area of the parabola, 
III, ,\ The area of the parabola is 256m 2 . 



n. 



MENSURATION. 211 

Prob. XXXVIII. To find the area of a parabolic frustum, 
having- given the double ordinates of its ends and the dis- 
tance between them. 

B z — b 3 

Formula. — A=%aX-^ r^- m which ah the distance be- 

-o** — o 4 

tween the double ordinates, B the greater and b the lesser double 

ordinate. 

Rule. — Divide the difference of the cubes of the two ends by the 
difference of their squares and multiply the quotient by -J of the 
altitude. 

I. What is the area of a parabolic frustum whose greater base 
is 10 feet, lesser base 6 feet, and the altitude 4 feet? 

i?3 A3 -IQ3 A3 

By formula, A=%a X ^C=p=tX ±*J^r^= S2 % B * ft * 
(1. 10 ft.=the greater base, 

2. 6 ft.=the lesser base, and 

3. 4 ft.=the altitude. 

4. 784 cu. ft.=10 3 — 6 3 =the difference of the cubes of the 
two bases. 

5. 64 sq. ft.=10 2 — 6 2 =the difference of the squares of the 
two bases. 

6. 12^ ft.=784-M34=the quotient of the difference of the 
cubes by the difference of the squares. 

17. .*. f X(4Xl2i)=32f sq. ft.=the area of the frustum. 
III. .*. The area of the frustum is 32f sq. ft. 

3. HYPERBOLA. 

1 . a 2 y 2 — b 2 x 2 = — a 2 b 2 is the equation of the hyperbola referred 
to its axes in terms of its semi-axes. 

b 2 

2. y 2 = -[lax — x 2 ) is the equation of a hyperbola referred 

to its transverse axis and a tangent at the left hand vertex. 

3. p=— i is the polar equation of the hyperbola. 

1 — e cos u 

Having given any three of the four quantites a, b, x,y, the oth- 
er may be found by solving the rectangular equation with refer- 
ence to the required quantity. 

Prob. XXXIX. To find the length of any arc of an hyper- 
bola, beginning at the vertex. 

Formula.— s=\'dy 2 +dx * =, J ( ( a ^^J+^ y ) dx= 

/ _1_ a*x^_ 113 a±+ia 2 b * 1-1-3-3.5 

y K +1.2.3 b± 1.2.3.4.5 3 s y +1-2-3.4.5.6.7 



x + -\- 



212 FINKEL'S SOLUTION BOOK. 

Rule. — 1. Find the parameter by dividing the square of the 
conjugate diameter by the transverse dia?neter. 

2. To 19 times the transverse, add 21 ti?nes the -parameter of 
the axis, and ?nultiply the sum by the quotient of the abscissa di- 
vided by the transverse. 

3. To 9 times the transverse, add 21 times the parameter , and 
multiply the sum by the quotient of the abscissa divided by the 
transverse. 

4- To each of the products thus found, add 15 times the para- 
meter , and divide the former by the latter; then this quotient 
being multiplied by the ordinate will give the length, nearly. 

[Bonny castle's I?ulc.) 

Note. — A parameter is a double ordinate passing through the focus. 

I. In the the hyperbola DA C, the transverse diameter GA 
=80, the conjugate 1/1=00, the ordinate 13 C=i0, and the 
abscissa ^4.^=2.1637 ; what is the length of the arc DA C? 

By formula, DAC= 
0/11 1 a " x2 

s = 2xil +T^-b^~ 

1.1.3 a*+4a 2 l>' 
1.2.3.4.5 b* 

1,13.3.5 
1.2.3.4.5.6-7 

20.658. ; __ ~™~ 

FIG. 20. 

1. 45=2 0Z*-±-OA=2bz -ha=the parameter LK which, in 
the figure, should be drawn to the right of DC, to be 
consistent with the nature of the problem. 

2. 1520=19X80=19 times the transverse diameter. 

3. 945=21X45=21 times the parameter. 

4. 2465=sum of these two products. 

5. .02704=2. 1637-r-80=quotient of abscissa and transverse 
diameter. 

6. 2465x.02704=66.6536=sum of the products multiplied 
by the said quotient. Also, 

7. [(80X9)+(45X21)]X^^=(720+945)X.02704= 

45.0216. Whence 

8. (15x45+66.6536)-H15x45+45.0216)=741.6536-v- 
720.0216=1.03004. 

9. .-. 1.03004xl0=10.3004=length of the arc A C, nearly. 
III. .-. The length of the arc is 10.3004. 




II.S 



MENSURATION. 213 

Prob. XL. To find the area of an hyperbola,the transverse 
and conjugate axes and abscissa being- given. 



Formulae.— (a) A=2fydx =2- f x \x 2 — a 2 )idx= 

Vx /2 —a 2 —ab\og\—^ =#'/_- 

ab log L-^4— ] ; or, (3) .4=4*, | *_ _ _ 

i r * 2 i 2 _ _j_ r * 2 1 3 __ & 

Z.b.l\a 2 +x 2 \ 5.7.9 [« 2 +**J c - 



-X 

a 



Rule. — 1. To the product of the transverse diameter a?id 
abscissa, add \ of the square of the abscissa, and multiply the 
square root of the sum by 21. 

2. Add Jf. times the square root of the product of the transverse 
diameter and abscissa, to the product last found and divide the 
sum by 75. 

3. Divide Jf. times the product of the conjugate dia?neter and 
abscissa by the transverse diameter, and this last quotent multi- 
plied by the former will give the area required, nearly. — Bonny- 
castle's Rule. 

I. If, in the hyperbola DA C, the transverse axis A G is 30, the 
conjugate diameter HI, 18 and the abscissa AB , 10 ; what is the 
area of the hyperbola DA C ? 



By formula (a), A=x / y / —ab logT*^*^ g2 l=25/-- 

15X9 lo, e [ 2 ^±^M!] = 300-1351o gc [ 25 +^] =300 

—135 log e 3=300— 135X1.09861228=151.687343,/ "being found 
from the equation a 2 / 2 — b 2 x" 2 = — a 2 b\ in which fl=15, £=9 
and * =15+10=25. 

1. 1- 21^30x10+4 x 10 2 =2lV300+500-r-7=2lV371.42857 
=21 X19.272 =404.712, by the first part of the rule. 

2. 2. (4V30xl0+404.712H-75=(4xl7.3205+404.712)-f- 
75=6.3199, by the second part of the rule. 

3. 3. .'. 18X 3 1 ° X4 x 6.3199=151.6776, by the third part 
of the rule, =the area of the hyperbola, nearly. 

III. /. 151.6776=the area of the hyperbola. 



214 FINKEL'S SOLUTION BOOK. 

Prol>. XLI, To find the area of a zone of an hyperbola. 

Formula.— A=2- f"tx 2 —a 2 )*dx 

=t x sl^^--ab log, P 1+ s/x l— a2 ~\__ b x J^=a~* _L 
a 2 2 L_ a J a l 1 ' 



ge L a — - y= x iy<i— x iyi— ab l °ge l a — -J + 

ab log, L ^i- - J-**.-**!— * log, L_-^^J, 

in which (x 2 ,y 2 ), and (*], jKx ) are the co-ordinates of the points 
C and L respectively. 

I. What is the area of a zone of an hyperbola whose transverse 
diameter is 2tf=10 feet, conjugate diameter 23=6 feet, the lesser 
double ordinate of the zone being 8 feet and the greater 12 feet? 

By formula, A= x^-x^-aHo^ \ _— ^ 2 , | 

But from the equation, when jy=)/ 2 =6, #=# 2 =10\/6 and when 
j/==y 1= =4, A'=x 1 =13^. Substituting these values of x 2 andjy 2 , 

we have ^=50^6— 66|— 30 iog.(^^° ) 

= \ 50V6— 66|—30 log,[f( v / 6-f-l)] \ sq. ft. 

Prob. XLiIl. To find the area of a sector of an hyperbola, 
RALO. 



For inula. — A=ab 



"*(W> 



Rule. — Find the area of the segment A KL by Prob. XL., and 
subtract it from the area oj the triangle KOL. 

I. What is the area of the sector OAL (Fig. 20) if OA=-a=5, 
OJ=b=$, and ZF=y=4 ? 



By formula, A=\ab log,(j+£)=7i log a (**+|)=7 
log«[ T 1 - Jf (3* + 20)]. But when ^=4, *=13£. Hence, 



IX. 



MENSURATION. 
HIGHER PLANE CURVES. 



215 



1. Higher JPlane Curves are loci whose equations are 
above the second degree, or which involve transcendental func- 
tions, i. e., a function whose degree is infinite. 

I. THE CISSOID OF DIOCLES. 

1. The Cissoid of Diodes is the curve generated by the 
vertex of a parabola rolling on an equal parabola. 

2. If pairs of equal ordinates be drawn to the diameter of a 
circle, and through one extremity of this diameter and the point 
in the circumference through which one of the ordinates is let 
fall, a line be drawn, the locus of the intersection of this line and 
the equal ordinate, or that ordinate produced is the Cissoid 
of Diodes. 

3. y 2 ==T- is the equation of the cissoid referred to rectan- 

- / %a — x 

gular axes. 

p=2a s'md tan# is the polar equation of the curve. 

Prob. XLIII. To find the length of an arc O AP of the 
cissoid. 



Formula.— s= OAP-. 



— ,)X 
X 



/* I 8tf— :># , ( \Sa— 3 

J ^*(2a — x) 6 ( N 2a — t 

—2+3 log, 

r ^ Vg(V3+2) -j) 

I. What is the length of the arc 
OAJV, in which case x=a? 

By formula, s=a i V5— 




Fig. 21. 



216 FINKEL'S SOLUTION BOOK. 

Prob. XLIV. To find the area included between the curve 
and its asymptote, BM. 

Formula.— A=2 j y dx=2 j N o *_ ^ x== l_ — i 

^2a x—\ 2a 

»J x ,\/2a — x(a-\-x)— 3 sin -1 • ,— \ =3tt« 2 , i. e., 3 times the area 

of the circle, OEB. 

Note. — The name Cissoid is from the Greek Hl6 6 0£idr}8,y like ivy, from 
Hl(T(?OS, ivy, €idoS, form. The curve was invented by the Greek geometer 
Diocles, A D. 500, for the purpose of solving two celebrated problems of 
the higher geometry; viz., to trisect a plane angle, and to construct two 
geometrical means hetween two given straight lines. The construction of 
two geometrical means between two given straight lines is effected by the 
cissoid. Thus in the figure of the cissoid, ED and OG are the two 
geometrical means between the straight lines OD and PG: that is, 
OD\ED::OG:PG. The trisection of a plane angle is effected by the con- 
choid. The duplication of the cube, i. e., to find the edge of a cube whose 
volume shall be twice that of a cube whose volume is given, may be effected 
by the cissoid. Thus, on KC lay off CH=2BC, and draw BH. Let fall 
from the point F, where BH cuts the curve, the perpendicular FR. Then 
RF=2BR. Now a cube described on RF is twice one described on OR', 

OR* OR 3 
for, since FR—y,OR=x, and BR=2a—x, we have RF* ——--——-—-, or 

x> 1 1 \r R 

\RF 3 —OR 3 . .'. FR 3 —20R 3 O^. E. D. 

2. THE CONCHOID OF NICOMEDES. 

1. The Conchoid is the locus formed by measuring, on a 
line which revolves about a fixed point without a given fixed 
line, a constant length in either direction from the point where it 
intersects the given fixed line. 

2. * x 2 y 2 =(b-\-y) 2 (a 2 — y*) is the equation of the conchoid re- 
ferred to rectangular axes. 




FIG. 22. 

3. p=b sec 6±ais the polar equation referred to polar co-ordi- 
nates. In this equation, 6 is the angle PO makes with AO. 

Prob. XL.V. To find the length of an arc of the conchoid. 
Formula.— *= A] 1+ (^Q ' 0<*=jVl+tan*0 8ec««tf. 



MENSURATION. 



217 



Prob. XX; VI. To find the whole area of the cinchoid be- 
tween two radiants each making an angle 6 with OA. 

Formula.— A=2Jir2dd=6* J (sec 6+ a) 2 dd= 

b 2 tan 0+2tf 2 64-33V« 2 — b 2 or b 2 tan d+2a 2 d, according as a is or 
is not greater than b. 1. The area above the directrix m m' and 

the same radiants =2#3 log tan ( j -(- - )-\-a 2 d. 

2. The area of the loop which exists when a is ^> b is a 2 



cos * 2ab log 

a 



-b* 



+6\'a- 



■b 2 



Ib—sjai—b* 
Note. — The name conchoid is from the Greek, KOyXosiSrfS,, 



from 
XOyX??, shell, and tidoe], form, and signifies shell-form. It was invented 
by the Greek geometer Nicomedes, about A. D. 100 for the purpose of tri- 
secting any plane angle. The trisection of an angle may be accomplished 
by this curve as follows: Let A OH be any angle to be trisected. From any 
point, G, in one side let fall a perpendicular, GF S upon the other. Take 
AF=i2GO, and with O as the fixed point, m m' as the fixed line and Pi 3 as the 
revolving line of which PD=.a is constant, construct the arc of the con- 
choid, PA II. Erect BG perpendicular to mm' and draw BO. Then is 
BOA one third of HO A. For bisect BK at /, and draw GI. Also draw IL 
parallel to G K. Since B 1—1 K, BL=LG and GI=B/=IK-=GO. By 
reason of the isosceles triangle BIG, we have the angle G/<9=2/ GBO— 
21BOA. But l_GIO—LIOG. ■'. 2lIOA=llOG, or IOA— %IH0A. 

Q^ E. F. 

3. THE OVAL OF CASSINI OR CASSINIAN. 

1. The Oval of Cassini is the locus of the vertex of the 
triangle whose base is 2a and the product of the other sides =m ii . 



4a 2 x 2 =m 4 is the tec I angular equation of the curve, in which 
2a=AB. 

3. r 4 — 2a 2 r 2 eos2#-|-0 2 — m 4 =0 is the polar equation of the 
curve. 

Discussion. — If a be ^> m, 
there are two ovals, as shown 
in the figure. In that case, the 
last equation shows that if 
OP P / meets the curve in P 
and P\ we have OP. OP'= 
Vtf 4 — m 4 \ and therefore the 
curve is its own inverse with 
respect to a circle of radius= 

VV— ZK 4 . 

FIG. 23. 




218 



FINKEL'S SOLUTION BOOK. 



4. LEMNISCATE OF BERNOUILLI. 

1. This curve is what acassznian becomes when m=a. 
above equations then reduce to 

2. (a^-f-y* ) 2 =2a^(x~ 2 —y^) and 

3. ^=20* cos 20. 

Prob. XLVXI. To find the 
length of the arc of the Leninis- 
cate. 

Formula. — s=J ^vjl+J ~tq j dd 



The 




dd= 



I 



FIG. 24. 

a 2 dr 



dr= 



_l_l_i i I. 



\ai 



3 1 
4'9 



/ a r~ 1 r 4 r 8 r 1 2 "1 

[^+i.- 6 +i-i^+i.i.t^+&c] 

+i-M-r3+ &c ~| = arc BPA. :. The entire length of the curve is 

441+U+M-H-&C] 
Prob. XJLVIII. To find the area of the lemniscate. 

Formula —A=4: f±r*d0=4:az C^ cos 20 d0= 
V2a* sin2#~] 4 '=2tf 2 . 

5. THE VERSIERA OR WITCH OF AGNESI. 

1. The Versiera is the locus of the extremity of an ordi- 
nate to a circle, produced until the produced ordinate is to the 
ordinate itself, as the diameter of a circle is to one of the seg- 
ments into which the ordinate divides the diameter, these seg- 
ments being all taken on the same side. 

2. Let P be any point of the curve, PD=y, 
the point P and 
OD=x, the abscis- 
sa. Then, by defini- 
tion, EP :EF.: 
AO-.FO, or x: 
EF: :2a \y. But 
E F=sJaEx FO 
=Y(2<2 — y)y. 
.-. x :V(2« — y)y : : 

2a :y. Whence x 2 y= FIG. 25. 

A.a 2 (2a — y) is the equation referred to rectangular co-ordinates. 



the ordinate of 




MENSURATION. 



219 



3. r(r 2 — r 2 sin 2 r9-j-4a: 2 ) sint3=8# 3 is the polar equation of 
the curve. 

Prob. XLIX. To find the length of an arc of the Versiera. 
Formula.-s=f^l + (^yd x =f4i +(x J^ ?lz dx. 

This can be integrated by series and the result obtained ap- 
proximately. 

Prob. L. To find the area between the curve and its 
asymptote. 



Formula 



.-A=2f; 



ydx=2XSa 



JC2a 
-. 
o X- 



dx 



+4tf : 



2^4« 2 tan^l °=4^ 2 , 



Rule. — Multiply the area of the given circle by 4- 

Note. — This curve was invented by an Italian ladv, Dona Maria Agnesi, 
1748. 

6. THE LIMACON. 

1. The Limacon is the locus of a point P on the radius 
vector OP, of a circle OPE from a fixed 
point, 0,on the circle and at a constant 
distance from either side of the circle. 

2. ( x *+y*—ax)* = &*( x *+y*) is 

the rectangular equation of the curve. 

3. r=a cos6±b is the polar equa- 
tion. In these equations, a= OA and \1 
b=PP. 

Prob. LI. To find the length of 
an arc of the Limacon. 



Formula. — s-- 




FIG. 26. 



Jsj(a cosd+b) 2 -\-a 2 sin*d dd=J^(a 2 -\-b 2 +2ab cos (,)d6= 

J \ j (a+b) 2 cos 2 - -+(a—b) 2 sin- r? I dd . .-. The rectification 

of the Limacon depends on that of an ellipse whose semi-axes are 
(a-\-b) and (a — b.) 

When a=b, the curve is the cardioid, the polar equation of 

which is r=a(l-{-cos ), and s= J ^^ 2 +( -je J dV= 



220 



FINKEL'S SOLUTION BOOK. 



a ftj2+2cos0 d6=±2aJcos±6 d0=2a H cos | Odd— 

2a I cos ^ 6 d&=8a=the entire length of the cardioid. 

Prob. MI. To find the area of the Limacon. 

Formula.— A= j V 2 d0=\ T^ (a cos 6+b)*d6= 

7r(^a 2 -\-6 2 ). When a=b, the curve becomes a cardioid, and A= 
\rta 2 . When d^>b, the curve has two loops and is that in the figure. 
r=a cos 0-\-b is the polar equation of the outer loop, and r=a 
cos 6 — b is the polar equation of the inner loop. The area of the 

/•C08- - 

inner loop is A=j\r 2 db=\\ < '(acosQ—b) 2 d6= 

(ia 2 +b 2 ) cos- x a—\b*la*—b' 1 . 

Note. — This curve was invented by Blaise Pascal in 1643. When a— 
2b, the curve is called the Trisectrix. 

7. THE QUADRATRIX. 

1. The Quadratrix is the locus of the intersection, P, of 
the radius. OD, and the ordinate 
QN, when these move uniformally, 
so that ON\OA\\LBOD\\n. 

fa — x n\ . 

2. y=xtanl .- 1 is the 

rectangular equation of the curve, 
in which a—OA, x=OJV, and 
y=IN. 

3. The curve effects the quad- 
rature of the circle, for OC.OB:: 
Off :arc ADB. 

Prob. Mil. To find the area 
enclosed above the x— axis. 

Formula. — A= fydx= 

/' fa — x tc^\ 
x tani .- jdx—4a 2 7r ~ l log 2. 

Note. — This curve was invented by Dinostratus, in 370 B. C. 

S. THE CATENARY. 

1. The Catenary is the line which a perfectly flexible chain 
assumes when its ends are fastened at two points as B and C in 
the figure. 




FIG, 27. 



MENSURATION. 22? 

2. y=\a(ea-\-e~Tx) is the rectangular equation of the curve, in 
which a—OA. A is the origin of co-ordinates. BA C is the 
catenary. M'AEM is the 
evolute of the catenary 
and is called the Trac- 
trix. To find the equa- 
tion of the curve, let A be 
the origin of co-ordinates. 
Let s denote the length 
of any arc AE ; then, if p 
be the weight of a unit 
of length of the chain, 
the verticle tension at E , 
is sp. Let the horizontal FIG. 28. 

tension at E ', be ap, the weight of a units of length of the chain. 
Let EG be a tangent at E, then, if EG represents the tension of 
the chain at E, EE and GF will represent respectively its hori- 
zontal and its vertical tension at B. 




dy FG_sp__s s __\ds 2 —dx 2 r f& 

dx EE ap a ' a~~ dx ~° J ^a 2 -\-i 



r as 

. .'. X=<7 I I = 

J \la z -\-s 2 
a\og (s-\-^a 2 -\-s 2 )-\-c. Since x=o, when s=o, c= — a log a. 

.-, x=a\ogl — h'Njl + i")- From this equation, we find s= 

-( ea-\-e a j which is the length of the curve measured from A. 

_ dy s dy s f * , x \ 
dx a dx a T V* ' e J 

Prob. LIV. To find the area of the Catenary. 

Formula. — A=fydx=f^a < e *+e*"\ ^*= 

(- --\ i 
e<i-\~e a J=a\y s — a 2 . This is the area included between the 

axis of x, the curve and the two ordinates, y 1 =a, y 2 =y. 

Note. — The form of equilibrium of a flexible chain was first investigated 
by Galileo, who pronounced the curve to be a parabola. His error was de 
tected experimentally, in 1669, by Joachim Jungius, a German geometer; 
but the true form of the Catenary was obtained by James Bernouilli, in 1691. 



2 FINKEL'S SOLUTION BOOK. 

9. THE TRACTRIX. 
1. The TractridC is the involute of the Catenary. 



2. x=a log \ a-\-V(a 2 — y 2 ) ! — a logj/ — \(a 2 — y 2 , is the rect- 
angular equation of the curve. 
Prob. LV. To find the length of an arc of the tractrix. 

Formula. — s= a log I - Y 

Proh. LVI. To find the area included by the four branches. 

ra 

Formula. — ^1= Cydx= — 4 / \la 2 —y 2 dy=na 2 . 

10. THE SYNTRACTRIX. 

1. The Syntraetrix is the locus of a point, Q, on the tan- 
gent, P T, of the Tractrix. 



2. 



x— a log \ c-\-*I(c 2 — y 2 ) !• — a logj — V(<: 2 — jk 2 ) is the rect- 
angular equation of the Syntraetrix, in which <; is QT, a con- 
stant length. 

11. ROULETTES. 

1. ^1 Roulette is the focus of a point rigidly connected 
with a curve which rolls upon a fixed right line or curve. 

(a) CYCLOIDS. 

1. The Cycloid is the roulette generated by a point in the 
circumference of a circle which rolls upon a right line. 

2. A Frolate Cycloid is the roulette generated by a point 
without the circumference of a circle which rolls upon a right 
line. 

3. A Curtate Cycloid is the roulette generated by a point 
within the circumference of circle which rolls upon a right line. 

4. x=versin~ 1 y — \2ry — y 2 is the rectangular equation of the 
cycloid referred to its base and a perpendicular at the left hand 
vertex. To produce this equation, let AJV=x and PA T =y, P be- 
ing any point of the curve. Let 0C= 
r=the radius of the generatrix OPL. 
Now AN=A 0—NO. But by con- 
struction A 0=urc P 0=versin " l FO y 
or versiri ~ x y to a radius r. JVO= 
PP=^PLx~PO==^\ f y( 2r—y ' = 
\2ry — y 2 . .*. x=versin~ 1 y — \2ry==y 2 
Or, we may have x=a(6 — sin6), and FIG. 29. 
y=a(l=cos 6) in which 6 is the angle, PCO, through which the 
generatrix has rolled. 




MENSURATION. 223 

For x=AO— NO. But AO=alPCO=ad, and NO=PF 
= PC sin IP CF=a sind. .-. x=ad — a sind=a(d — siitO) , y= 
O C— CF=a— CF. But CF=P C cos Z P CF=a cos 6 . • . y= 
a — a cosd=a(l — cos 6). 

Prob. LTVII. To find the length of an arc of the cycloid. 
Formula.— s=j ^Jl+( y- ) dy=J A l +^~ — i d y= 

* Vrfy J *■' Iry — y 

^2rf(2r—yy i dy=—2^2r(2r—y) h ^-c. Reckoning the arc from 
the origin, c=4r\ and the corrected integral is s= — 2(2r) 

(2r— y) +4/\ Wheny=2r, s=4r. .-. The whole length of the 
cycloid is 8r=4tZ),t e., the length of the cycloid is 4 times the 
diameter of the generating circle. 

Rule. — (1) Multiply the corresponding chord of the genera' 
trix by 2. To Jind the length of the cycloid : 

(2) Multiply the diameter of the generati?ig circle by 4- 

I. Through what distance will a rivet in the tire of a 3-ft. 
buggy wheel pass in three revolutions of the wheel? 
By formula, *=3(8r)=24xlift.=36ft. 
r 1. 3ft.=the diameter of the wheel. Then 
I 2. 12ft=4X 3 ft.=di stance through which it moves in 1 
II. < revolution. 

I 3. .'. 36ft. =3X 12ft. ^distance through which it moves in 
3 revolutions. 
III. .*. It will will move through a distance of 36ft 

Prob. LVTTT. To find the area of a cycloid. 
Formula.— A=2 fydx=2 I y " y = 

J 'o \-2ry—y 2 

Sr 2 vcrsin- 1 2=S7tr 2 . 

Rule. — Multiply the area of the generating circle by 3. 

I. What is the area of a cycloid generated by a circle whose 
radius is 2ft.? 

By formula,^4=37rr 2 =3^2 2 =12^=37.6992 sq. ft. 
( 1. 2ft.=the radius of the generating circle. 

II. < 2. 7r2 2 =12.5664 sq. ft. -—the area ot the generating circle. 
( 3. 37r2 2 =37.6992 sq. ft.=the area of the cycloid, 

III. .-. The area of the cycloid is 37-6992 sq. ft. 

Prob. LIX. Awheel whose radios is r rolls along* a hori- 
zontal line with a velocity v'; required the velocity of any 
point, P, in its circumference; also the velocity of P horizon- 
tally and vertically. 



224 FINKEL'S SOLUTION BOOK. 

Since a point in the circumference of a wheel describes, in 
space, a cycloid, let P ', Fig. 29, be the point, referred to the 
axes AA / and a perpendicular at A. Let (xy) be the coordi- 
nates of the point; then will the horizontal and vertical 
velocities of P be the rates of change of x and y respectively. 

y 

O being the point of contact, A 0=r versin' 1 -. Since the cen- 

r 

ter C, is vertically over O, its velocity is equal to the rate of in- 
crease of A 0. In an element of time, dt, the center C will move 

the distance dl r versin"- f — /- — ~ . .*. Its velocity v' = 

V rJ \2ry—y 2 J 

the distance it moves divided by the time it moves, or v'= 

rdy _ r dy dy \2ry — y 2 . . 

J -±-dt— — -f-. .-. — Jjt = — - — ±—v'= the velocity 

V2ry— ; y 2 \2ry—y 2 dt dt r 

vertically. . . . (1). 



From the equation of the cycloid, x=r versin 1 <~- — \2ry — y 2 , we 

y 

have dx=-= dy. Now dx-Z-dt=the velocity of the point 

\2ry — y 2 

horizontallv. But dx-^-dt, or— -,— -^-. Substituting the 

dt s/2ry—y' z dt 

value of -v, we have — = - v' (2). An element of the 

dt dt r 

curve APBA / is ds and this is the distance the point travels 

ds 
in an element of time, dt. .*. — = the velocity of the point, P. 

at 

But ds=V ' dy 2 -\-dx 2 =^\( ^ ry ~ y +£?) v / dt=^\-2-v / dt i since, 
from (1), dy= fo^T? 2 ^ and, from (2), dx=(%Q. /. By 

dividing by oft, we have S=v=^\— v / =^ the velocity of the 

point, P (3). From (1), (2), and (3), we have, 

ds 
di~~ w ' aT~ ~dt 



if^.^;^,and£=^ 






MENSURATION. 225 

Hence, when a point of the circumference is in contact with 
the line, its velocity is 0\ when it is in the same horizontal plane 
as the center, its velocity horizontally and verically is the same 
as the velocity of the center, and when it is at the highest point, 
its motion is entirely horizontal, and its velocity is twice that of 

ds \ly ^2ry 

the center. Since -— = ->.— 1> / =- — -v / , we have by proportion, 
dt N r r ♦ 

^: v'\ : V^r~y\r. But V 1 b r y=y/(PF' l +FO' l )=PO. 

.'. The velocity of P is to that of C as the chord P O is to the 
radius CO; that is, P and C are momentarily moving about O 
with equal angular velocity. 

■•-■■ (b) THE PROLATE AND CURTATE CYCLOID. 

1. x=d(V — m sin#), y=a(l — mcosd) are the equations in 
every case. 

2. The cycloid is prolate when m is >1 as AIP'BTA', Fig. 
30, and curtate when m is <C1, as PP. These equations are found 
thus: Let CP=ma, and lOCP=d. Then x=AJV=A O— ON. 
But ^4 0=arc subtended by lOCP=aQ 1 and ON=PtXsin 
lNPC=mas\n LNP C ( = IP CL=n—V) =ma 6 in (tt— #)= 
ma s'mO. .'*. x=ati — ??za sin 6=a(d — m sin#). y=PN=OC-\- 
PC coslNPC (=IPCL=^7T— 6) =a+ma cos (n— 6) =a— 
ma cos# =a( 1 — m cos g ). The same reasoning applies when we 
assume the point to be P' . 

Note. — These curves are also called Trochoids. 

Prob. L.X. To find the length of a Trochoid. 

Formula.— s= J si d x i+dy' 1 . 
Since x=a(8 — m sin#), dx=a(l — m cos P)d6 ; and since y= 

a(l—m cosd ), dy=amsinftdd. .'. s= Ndx~*+dy 2 ==a / \][" (1— ■ 

)2+m 2 sin* 6~\d6=a f^ ^\(l-\-m 2 — 2m cos g)d6= 

Aa f**«\ \ (l+w)2_4^cos 2 cp I dcp=4a(l+m) /^"^(l— 
cos 2 cp)dcp. 



m cos 



(l+m) 



I. If a fly is on the spoke of a carriage wheel 5 feet in diame- 
ter, 6 inches up from the ground, through what distance will the 




226 FINKEL'S SOLUTION BOOK. 

fly move while the wheel makes one revolution on a level plane? 

Let Cbe the center* of 
the wheel, in the figure, 
and P the position of the 
11 v at any time. Let OC= 
the radius of the carriage 
wheel =a=2\ ft, PC= 
2 ft, and the an^le OCP 
=6. Let (x,y) be the co- 
ordinates of the point P. 
Let T 7 , a point at the inter- #& 30. 

section of the curve and Albe the position of the fly when the 
motion of the wheel commenced. Then since x—a(ft — msir\6) 
and y=a(l — mcos6), we have dx—a(l-^-m cos 6)d6, and dr- 
am sinOdd. .-. s=J\l(dx*+dy* )=£ 'J. ]«*.( 1— m cos#) 2 + 

a 2 w 2 sin 2 j d9=ajyjlJ r m2— 2 w cos 6)d0=ia f* [*\ \ (1+ 
w) 8 — 4wcos 2 <p f d<p, in which <p=|#. But PC=2 ft, and 
since / > C=^a=^x2| ft, ^= 2H-2i=f. .-. .y==4x2i / 
-s| | (1+|) 2 — 4x|cos 2 ^ | d<p=10 [^^($1— 80 cos* cp)d<p=; 
* is/^a-l! cos 2 <p)d<p=$n | l_|o_3( T .V) 2 (M) 2 - 

5( T 1 i3) 2 (M) 3 -7( 1 1 «^) 3 (lf ) 4 ~&c | =18.84 ft 
II. .•: The fly will move 18.84 ft. 

Prob. IiXI. To find the area contained between the tro- 
choid and its axis. 

Formula. — A= fydx=2a 2 I ( 1 — m cos 6) ( 1 — 

m cos 0)de=2a 2 f 7V (l—m cos 6yd6=2a 2 (^6— 2m sintf+A/w 2 
(6— sin 0cos#)V=2r7 2 (n+\m 2 n). When «i=l,' 



the curve is 
the cycloid and the area=37T« 2 as it should be. 

* When <pis replaced by (\7t-^-cp), this is an elliptic integral of 
the second kind and may be written 4«^(|j,^). 



MENSURATION. 



227 



(c)-EPITROCHOID AND HYPOTROCHOID. 

1. An Epitrochoid is the roulette formed by a circle roll- 
ing upon the convex circumference of a fixed circle, and carrying 
a generating point either within or without the rolling circle. 

2. An HypOtrochoid is the roulette formed by a circle 
rolling upon the concave circumference of a fixed circle, and 
carrying a generating point either within or without the rolling 
circle. 



3. #=:(<Z-j-#)C0S 



ibcos 



«+* 



y=(a-\-b) sin 6- 



cos— j—tr 
b 



are the equations of the epitrochoids. 

In the figure, let C be the center of the fixed circle and O the 
center of the rolling circle. Let FP'Q be a portion of the curve 
generated by the point P' situated within the rolling circle, and 
let CG=x and P' G—y be the co-ordinates of the point, P / . 

Let A be the position of P when the rolling commences, and 
q)=/_PQC through which it rolled. Draw OA'perpendicular 
to CGand P'l perpendicular to OK\ draw DP and DP'. Let 
OP / =mOP=mb and the angte A CD=6. Then x=CG=CK 
-yKG=KC+P'L But CK—OC cos ti=(a+6) cos 6 andP'I= 
P'OcoslOP'/.-mbcos 

\ n — ( (p+\-W ) \ = — mb 

,co *(<p-\- #)... But arc 
AD=arcPD. .: ad= 

bqj. Whence cp=~d. 

.-. P'/=— m5cos a -^e 9 

b 

and x—(a-\-6) cos 6 — 

mlcos^d. v=P'G 

b J 

=.IK=OK—OL But 

OK=OCs\nlKCO= FIQ. 31. 




(ff+J)sin0, and OI=OP / sin/ OP / /=wb sin] n— {(p+d) \ = 



mb sin(<^-|-#)=/;^sin-^ — 6. 



'• J'=(tf+£)sin 8 — ?nb sin J 0. 



b ' - ' ''--- b 

If m—1, the point P' will be on the circumference of the roll- 
ing circle and will describe the curve APJV which is called the 



228 FINKEL'S SOLUTION BOOK. 

Epicycloid The equations for the Epicycloid are x=(a-\-b)cosd— 

b cos ~T- 0i andy=(a-\-b) sin 6 — b sin — f- 0. The equations forthe 

Hypotrochoid may be obtained by changing the signs of b and mb, 
in the equations for the Epitrochoid. .*. x—(a — 3)cos 0-\-mb cos 

— - — #, andy=(a — £)sin Q — 7?zb sm — — 6 are the equations for the 
o o 

Hypotrochoid. If /w=l, the generating point is in the circumfer- 
ence of the rolling circle and the curve generated will be a Hypo- 
cycloid. .'. x=(a — b) cos# -{-b cos — y— 0, and y=(a — b)sin — 

b sin — — 9 are the equations of the Hypocycloid . 

Prob. IjXII. To find the length of the arc of an epitro- 
choid. 

Formula.— s=Jsidx*+dy*=j ^ ^T—( a +b) sin 6+ 
m (a+b) sin^±^T -\-^( a +b)cos 0— m (a+b) cos^t^l * | dd 
=(a+b)f I | 1+m 2 — 2m(smd sin^ 6+ 

cos0cos0^±^) \dd=(a+b)J^l-\-m*-~2mcosjd)d6. This 

may be expressed as an elliptic integral, E(k, q>), of the second 

2b 
.kind, by substituting (zr | q)) for 6, and then reducing. 

2. By making m=l, we have s=(a-\-b)^2 /V(l — 
cos- 0) dQ, the length of the arc of an hypocycloid. 

3. By changing sign of b, the above formula reduces to t= 
(a—b) fv / (T+w 2 +2»zcos-2 6 )de, which is the length of the arc 

of an hypotrochoid. 

4. By making m=l, in the last formula, we have s= 

a i 
(a — b)\ f 2 f(l-}-cos - oy*dd, which is the length of the arc of -<\\ 

hypocycloid. 

I. A circle 2 ft. in diameter rolls upon the convex circum 
Terence of a circle whose diameter is 6 feet What is the length 



MENSURATION. 229 

of the curve described by a point 4 inches from the center of the 
rolling circle, the rolling circle having made a complete revolu- 
tion about the fixed circle ? 

In Fig. 31, let Obe the center of the rolling circle; C the center 
of the fixed circle ; C/}=3 ft.=a, the radius of fixed circle; OD= 
1 ft.=£, the radius of the rolling circle; OP=4t inches=^ of 12 
\nchcs=mb the distance of the point from the center; and P the 
position of the point at any time after the rolling begins. Let 0= 
the angle A CI) and cp=the angle POD through which theroll- 
ingcircle has rolled. Then we have, as previously shown, the 
equations of the locus P, 

x=(a-\-b)cos 6 — mb cos(cp-\-0)=(a-\-b) cos 6 — mb cos ~-^~ ft, 

y=(a-\-b) sin — mb sin (<p+0 )=(#-{-£) sin — mb sin -~ — 0. 

From these equations, we can find dx and dy. 

.-. By formula, s=fs/dx 2 -\-dy 2 =6(a+b)J \^"l+w 2 — 
2m cos j 0)d 0=24 /^Vfl-H*) 8 — J cosS9)d0=S f*V(iO— 
6 cos Z0)d0. Let 3 0=2$. Then s=S /"^VflO— 6 cos 3 0)d6= 
21^^(1— fcos^)<#> =21ijf l,r [l— ffcos^— i.i.(|) 2 
cos 4 ^— i-i.f (i) 8 cos«#— f i.ff(f) 4 cos 8 ^— &c.~|<ty= 
21i | ^— i-l[i(isin2^+^)]-i.i.(|) 2 [i(isin4^+2sin2^+30)] 
-ii-i(l) 8 [^(isin6^+!- S in4^+Vsin2^+10^)]-&c. | *', = 



101* | l-(i) 2 (f )-Ki:f ) 2 (l) 2 -i(i.|.f ) 2 (f) 3 -Kf l-f-J) 2 

(|)4_&c. | =10$* x. 773=26.9 ft., nearly. 

Remark. — When the point is on the circumference of the roll- 
ing wheel, the length of the curve generated by the point is s= 

(a+ b) jV(l+*/ 2 — 2™ cos? 0)d6=(a+b) Js/'(i+l— 2 co&j0)d0. 
If we let the conditions of the above problem remain the same, 



230 FINKEL'S SOLUTION BOOK. 

only changing the generating point to the circumference, we have 
for the length of the curve, j=6V2(3+1) / 5 V(i-"Cos30)fl?0== 

48/ V(l — \ cos 2 (p)dcp, where q>=\ H. Expanding this by the 

Binomial Theorem and integrating each term separately, s= 

24* j i_(*)t( i )_tfi.|)«tt)._Ki.i. t )»Ci)«-&c. 

I. A circle whose radius is 1 ft. is rolled on the concave cir- 
cumference of a circle whose radius is 4 ft. What is the length 
of the curve generated by a point in the circumference of the 
rolling circle, the rolling circle having returned to the point of 
starting? 

x=(a — b) cos 6-\-b cos d y 

y=(a — b) sin — b sin ^-j-6, are the equations of the curve 

which is a hypocycloid. In these equations a=4 and £==1. 
... #=3 cos #-|-cos 3#=4 cos 3 /9, and 
^=3sin# — sin 3 #=4 sin 3 6. Whence, 



cos 



:bs0=4J(0' and sin 0=^0 
... cos 2 0+sin 2 0=Q^*+QCy. But cos* 0+sin" 0=1. 

••• (0 s +(9 Wm«i, 

#§_|-y§=4£, which is the rectangular equation of the curve, 
,\ By formula^=V(^ 2 -\-dy* )=4 /'Y*!±^iV dx= 

Aah f*x '*dx=4a* ff ^~| "=60=6 X4=24 it 

X. PLANE SPIRALS. 

1. A Plane Spiral is the locus of a point revolving about 
a fixed point and continually receding from it in such a manner 
that the radius vector is a function of the variable angle. Such 
a curve may cut a right line in an infinite number of points. 
This would render its rectilinear equation of an infinite degree. 
Hence, these loci are transcendental. 

2. The Measuring Circle is the circle whose radius is 
the radius vector of the spiral, at the end of one revolution of the 
generating point in the positive direction. 



Mensuration. 



231 



3. A Spire is the portion of the spiral generated by any- 
one revolution of the generating point. 

1. THE SPIRAL OF ARCHIMEDES. 

1. TTie Spiral of Archimedes is the locus of a point 
revolving about and receding from a fixed point so that the ratio 
of the radius vector to the angle through which it has moved 
from the polar axis, is constant. 

2. r=a6 is the polar equation of this curve. 

Prob. LXIII. To find 
the leng-th of the spiral 
of Archimedes. 

Formula . — s= 



dd- 



C s l(r' i +a*)dO=aC«(l+ 

£*k>gj#+V(i+tf 2 ) 

which is the length of the 
curve measured from the 
origin. 




FIG- 32. 



te<77rV[l+(27r) 2 ]+JKzlog 27r+ v / [l+(27r ) 2 ] is jhe length 

of the curve made by one revolution of the generating point. 
Prob. LXIV. To find the area of the spiral of Archimedes. 

Formula,— A=^Jr 2 de=ia 2 J6^d6=^a 2 e^=^-, the 

area measured from the origin. 

2. THE RECIPROCAL OR HYPERBOLIC SPIRAL. 

1. The Meciprocal or Hyperbolic Spiral is the locus 
of a point revolving around and receding from a fixed point so 
that the inverse ratio of the radius vector to the angle through 
which it has moved from the polar axis, is constant. 

a 

*■» 

6 

Prob. LXV. To find the length 
of the Hyperbolic Spiral. 

Formu la. — s= 



2. r=~ is the polar equation of the Hyperbolic Spiral. 



:fl V (l+0 2 )+logj <9+V(l + * 2 ) g- 




FIG. 33. 



232 FINKEL'S SOLUTION BOOK. 

^(l+^^log j +#V("l+# 2 ) (— 0- 1 */l+6 r , is the length of 
the spiral measured from the origin. 

Prob. LXVI. To find the area of the Hyperbolic Spiral. 

2 



/df) a* 

H~ 2 ~~20 1 the 



area 




measured from the origin. This result must be made positive 
since the radius vector revolves in the negative direction. 

3. THE LITUUS. 

1. The LitUUS is the locus of a point revolving around 
and receding from a fixed point so that the inverse ratio of the 
radius vector to the square root of the angle through which it has 
moved, is constant. 

a 

2. r = z Tfa 1S tne equation of the 

Lituus. 

Prob. LXVII. To find the leng th 
of the Lituus. 

FIG. 34. 

Formula.— s= f^ r2 +\Jg) *<*£fi*0"* jfvl l+4fl 2 )d.d= 

Prob. LXVIII. To find the area of the Lituus. 
Formula.— A=± j ^d0=~ fj- = ^ a2 log °' 
4. THE LOGARITHMIC SPIRAL. 

1. The Logarithmic Spiral is the locus generated by a 
point revolving around and receding from a fixed point in such 
a manner that the radius vector increases in a geometrical ratio, 
while the variable angle increases in an arithmetrical ratio. 

2. r=ad is the polar equation of the Logarithmic Spiral. If 
a is the base of a system of logarithms, this equation becomes 
/9=log r. 

Prob. LXIX. To find the length of the Logarithmic 
Spiral. _____ 




MENSURATION. 233. 

(m 2 -\-l)*dr=^~{nt*-\-l)r, where m is the modulus of the system 

of logarithms. 

Prob. LXX. To find the area of the 
Logarithmic Spiral. 

Formula.— A=%fr 2 d8=~- I rdr— 

\i?ir 2 . Since m=l, in the Naparian System 
of Logarithms, A—^r 2 , i. e., the area is \ of 
the square of the radius vector. 

FIG. 35. 

XI. MENSURATION OF SOLIDS. 

Prob. LXXI. To find the solidity of a cube, the length of 
its edge being" given. 

Formula. — V=( edge ) X ( edge ) X (edge )=( edge ) 3 . 

Rule. — Multiply the edge of the cube by itself, and that 
product agai?z by the edge. 

I. What is the volume of a cube whose edge is 5 feet? 

By formula, F=(edge) 3 =(5) 3 =125 cu, ft. 

T j { 1. 5 ft.=the edge of the cube. 

( 2. 5x5x5—125 cu. ft.=the volume of the cube. 

III. .*. The volume of the cube is 125 cu. ft. 

JRemark. — Some teachers of mathematics prefer to express the 
volume by saying 5x5X5X1 cu. ft. =125x1 cu. ft.=125 cu. ft. 

Prob. ixXII. To find the volume of a cube, having given 
its diagonal. 



Formula.— I 



<S) 



Utile. — Divide the diagorial by the square root of 3, and the 
cube of the quotie?it will be the volume of the cube. 

What is the, volume of a cube whose diagonal is 5L9615 inches? 



234 FINKEL'S SOLUTION BOOK. 

By formula, V^ *V =C^Y ^™*™y = 
y Vy^y V va J V1.73205V 

27,000 cu. in. 

fl. 51.9615 in— the diagonal. 

I|J 2. 30 in. =5 1 .00 1 5 in-T-V3=x:51.9615 in.-r-1.73205=the edge 
I of the cube. 
[3. .'. 30X30X30=27,000 eu. in.==the volume of the cube. 

III. /.The volume of the cube whose diagonal is. 51.9615 in., 
is 27,000 eu. in. 

Prob. L.XXII1. To find the volume of a cube whose surface 
is given. 

3 



Formula.— F===^>J- J 



Rule. — Divide the surface of the cube by 6 and extract the 
squae root of the quotient. This will give the edge of the cube. 
The cube of the edge will be the volu?ne of the cube. 

I. What is the volume of a cube whose surface is 294 square 
feet? 

By formula, F=(J|.y=(^^y s ^(Vi9)»==7»== 

243 cu. in. 

fl. 294 sq. ft.=the surface of the cube 
II 1 2. 49 sq. ft.=294 sq. ft.-j-6=area of one siae of the cube, 
I 3. Vi9=7 ft.— length of the edge of the cube. 
|^4. .*. 7X7X7=343 cu. ft.=volume of cube- 
Ill. .*. 343 cu. ft. is the volume of a cube whose surface is 
294 sq. ft. 

Prob. LXXIV. To find the solidity of a parallelopipedon. 

Formtda, — V—lXbXt<, where /=length, £=breadth, 
and /=thickness. 

Rule. — Multiply the length, breadth and thickness together. 

I. What is the volume of a parallelopipedon whose length is 
24 feet, breadth 8 feet, and thickness 5 feet? 
By formula, F=/x JX/=24 XSX5=960 cu. ft. 

fl. 24 ft.=the length. 
TT J 2. 8 ft.=the breadth, ami 
' j 3. 5 ft.=the thickness. 
U. .-• 24X8X5=960 cu. ft.=the volume. 
Ill, .'. 960 cu. ft.==thc length of the parallelopipedou. 



MENSURATION.; 285 

Prob. LXXV. To find the dimentions of a parallelopipe 
don, having given the ratio of its dimensions and the volume. 

Formitia.—Z=&[V-7-{7rtXnX!P)~\'m; b=^\V^ 

(mXnXfl)]n ; and £=$ / .[ Vrr-( mXn Xft)}fi, where m, n, and p 
are the ratios of the length, breadth, and thickness respectively. 

Rule. — Divide the volume of the parallelopipedon by the pro- 
duct of the ratios of the di??iensions, and extract the the cube root 
of the quotient. This gives th<, G. CD. of the three dimensions. 
Multiply the ratios of the dimensions by the G. CD., and the 
results will be the dimensions respetively. 

I. What are the dimensions of a parallelopipetlon whose 
length, breadth and thickness are in the ratios of 5, 4 and 3; 
and whose volume is 12960 cu. ft. ? 

By formula, /==yfl2960-S-(5x4x3)]5=30 ft.; ^^[12960-r- 
(5x4x3)]4=,24ft; and /^=^ / [129,60-r-(5x4x3)]3^18 ft. 

1. 5=the quotient obtained Jby dividing, the length by the 
G. C. D. of the three dimensions. ,, ', 

2. 4=the quotient obtained by dividing the breadth by the 
G. C. D.oP the three dimensions. \ 

3. 3=the quotient obtained by dividing the thickness by 
the G. C. D. of the three cjin^'ensions. 

4. .-. 5*G. C. D .=tiie length, 

5. 4XG. C. D.=the breadth, and 
IlA 6. 3XG. C. D.=the thickness. 

7. /. (5XG. C D.)X(4XG C.D.)X(3><G.C.O.)=:60X 

(G. C. D.) 8 — the volume of the parallelopipedon. 
■8. .-. 60(G. C. D.) 8 ^12960cu. ft: 
9. (G. C. D.) 3 =12960-r-60=216. 

10. .-. G. C.;D.— ^216^6. 

11. ,\ 5X(0. C D.)=5X6=30 ft.=the length, 

12. 4>X(G. C. D.)=4XB=24 ft— the breadth, and 
.13. 3X(G. C. D, ; )=3X6=1§ ft,=tbe thickness. 

III. '.-. 30 ft., 24 ft.; and 18' ft. are the dimensions of the par- 
allelopipedon. 

Prob. LXXVI. To find the convex surface of a prism. 

Formula. — S=pX&, hi which/ is the perimeter of the 
base and a the altitude. 

Rule. — Multiply the perimeter of the base by the altitude. 



FINKEL'S SOLUTION BOOK. 



I. What is the convex surface of the prism ABC — D, if the 
altitude AE is 12 feet, AB, 6 feet, A C, 5 feet, and BC, 4-feet.? 

By formula, S=«x/=12x( 9+5+4 )=180 sq. ft. 

( 1. 12 ft =the altitude of the prism. 

II. } 2. 6 ft.+5 ft.+4 ft.=15 ft.=the perimeter of the base. 

( 3. .'. 12X15=180 sq. ft.=the convex surface of the prism. 

III. .'. The convex surface of the pris'm is 180 sq. ft. 

Remark. — If the entire surface is required; to the convex sur- 
face, add the area of the two bases. 

Formula. — T—S-\-2A, where 2A is the area of the 
base, S the convex surface, and T the total surface. 

Prob. LXXVII. To find the volume of a prism. 

Formula. — V=aXA, where A is the area of the base, 
a, the altitude. 

Rule, — Multiply the area of the base by the altitude. 

I. What is the volume of the triangular prism ABC — D, 
whose length AE is 8 feet, and either of the equal sides AB, 
BC, or AC, 2i feet? 

By formula, F=aX^=8x[(2i) 7 i^]=l-i v S= 21 -6506 cu. ft 

1. 8 ft.==the altitude AE. 

2. 2\ ft.=the length of one of the equal 
sides of the base, as AB. 

II J 3. (2i)4V / 3==the area of the base ABC, 
by Prob. XI. 
4. .-. 8x(2|) 2 iV3=12v3=21.6506cu. ft. 
=the volume of the prism. 



III. 
prism. 



21.6506 cu. ft.=the volume of the 




FIG. 36. 



1. THE CYLINDER. 
Prob. LiXXVIII. To find the convex surface of a cylinder. 

Formula. — S==a X C, in which a is the altitude and C the 
circumference of the base. 

Rule. — Multiply the circumference of the base by the altitude. 

I. What is the convex surface of the right cylinder A GB — C, 
whose altitude EF is 20 feet and the diameter of its base AB is 
4 feet? 



II. J 




MENSURATION. 237 

By formula, S=aX C=20x 4^=80^=251.32736 sq. ft. 

(1. 20 ft.=the altitude EF. 

2. 4 ft.=the diameter AB of the base. 

3. 12.566368 ft.=4*=4x3.14i592=the 
circumference of the base. 

4 ,-. 20X12.566368=251.32736 sq. ft.= 
the convex surface of the cylinder. 

III. .'. The convex surface of the cylinder is 
251.32736 sq. ft. 

Remark. — If the entire surface is required; to 
th the convex surface, add the area of the two 
bases. 

Fomula.— T=S-\-2A=27raP+27rB 2 . FIG. 37. 

Prob. LXXIX. To find the solidity of a cylinder. 

Formula. — V—aXA, in which A is the area of the 
base. 

Rule. — Multiply the area of the base by the altitude. 

I What is the solidity of the cylinder A GB — C, whose alti- 
tude FE is 8 feet and diameter AB of the base 2 feet? 

By formula, V=a X ^4=8 x(l 2 7r)=87r=8X 3. 141592= 
25.132736 cu. ft. 

1. 8 ft.=the altitude, EF. 

2. 2 ft.^the diameter, AB, of the base. 

3. 3.141592 sq. ft.=7r7? 2 =7rl 2 =area of the base. 

4. .-. 8X3.141592=25.132736 cu. ft. 

III. .*. 25.132736 cu. ft. is the volume of the cylinder. 

2. CYLINDRIC UNGULAS. 

1. A Cylindric Ungula is any portion of a cylinder cut 
off by a plane. 

Prob. LiXXX. To find the convex surface of a cylindric 
ung-ula, when the cutting plane is parallel to the axis of the 
cylinder. 



IlJ 



.. t ■< ■ 



. S=aSds ^ a £^^-2ar sin- t=a X 

arc of the base. 

Rule. — Multiply the arc of the base by the altitude. 

I. What is the surface of the cylindric ungula API — Q, 
whose altitude AD is 32 feet and height A T of the arc of the 
base, 2 feet and cord PI of the base 12 feet? 



FIN KALI'S SOLUTION BOOK. 



;•■■■' ■■;;"! : ">r\v. I .'IS; --:Vv ;Hv=-. ■ r. ;.• '••.. j^-..- - * *> ■=< .. -" : / V \ 

By formula, S"= a'Xrrrc PA I— air siii" 1 ^ — #X2r ■?in~ , ( ^— j 

S2(5!±^^ sirr ^ i|^ 40 ^l|^^ ^411.84 ^. ft., neatiy. 

The arc corresponding to the sjnf, is found, from a tnlple of 



natural sines and cosines to be (36° 52^/-^-SQ0°) of %rt'bi 

1. 2 ft.=tne ! rVei^ht \#'5*of fhe arc PAP 

2. 12 ft.^= tne ; length of the chord >>>: 

3* 12.87 mm^ypmmi^ ' y - :% -^ 

■ , 6Qx^+33^2^ 7 thelengt r fthe 
arc 7M7, byProb. XXV. 
4 . .32X12,87=411.84 sq. ft = convex ( 
surface PA I—D. 

111. .*. The convex surface of the cylindric. 

ungula PAJ— Q is '411.84' sq. "ft" 

,:'JF-\ ■: '■ .'■' y '. ;.., ; v>b'.'' ■ x '• \) ' ■: ; - r -■;■-'=• -•■"'"' 
Remark,— r is found, bv Prob. XX, formula 



86400 



7t. 




FIG 38. 



Prob. LXXXI. To find the volume of a cylindric ungula, 
whose cutting- plane is parallel to the axis. , 

Formula.— V=2 f* T^^FdyJx.dz?- 2zy(r*—y 2 ) 



V'*-y 3 



)*=2tf f^i^—v 2 )dy— 



=2« C V C"' * dydx^ay^—y*) 
2ay{r^-y^) = a \y(r*-y* f+r* sin'* ^-2y(r^y* ) 2 \,= > 
■•*©|r 2 sin vj ^- T -^(,r 2 ^)/ 2 ,).vj i , in which ^/ is half the chord ' of the 

base., In this formula ( r 2 'sin~f : - 1 is the #rea of the sector 

APEIA, andj/(r 2 — -y 2 )* is the area of the triangle T^S 1 / formed 
by joining the center E with P and I. .. ..-• 

; 3^uje.-ri^/^/^ th$ area qf the base by the altitude, 

I. "Whatis the volume of the cylindric ungula PJA—D^ if 
/'/is 12 feet, ^4 7^2 feet, and'aititude AD 40 feet? 



MENSURATION. 239 

By formula, V=aA=a Jr 2 sin '?— y(r 2 — j/ 2 )*j =40} 10 2 sin '* 

-6(10 2 -6 2 )* [=4000 sin' 3_ 1920 :=400o(^^7r)^1920= 
2574.016- 1920=654.016 cu. ft. 

fl. 40 ft.=the altitude AD. 
2. 2 ft.=the height vl 7^ of the arc of the base. 
o\ 12 ft.=the chord PI of the base. 

02 

H.|4. 16£ sq. ft.=- — — +f of(2Xl2)=the area ot the base, 

by rule,Prob. XXVIII. 
5. .-. 40X16^=653^ cu. ft.=the volume of the cylindrical 
ungula PIA—D. 

III. .'. 653^ cu. ft.=the volume ot the cylindrical ungula. 

Remark. — A nearer result would have been obtained by finding 
the length of the arc PAI and multiplying it by half the radius. 
This would give the area of the sector IEPA. From t he area of 
the sector subtract the area of the triangle PIE formed by join- 
ing P and /with E y and the remainder would be the area of the 
segment PI A. 

Prob. LXXX1I. To find the convex surface of a cylindric 
ungula, when the plane passes obliquely through the op- 
posite sides of the cylinder. 

Formula. — S=\(a-\-a' )2nr, where a and a' are the 
least and greatest lengths of the ungula and 2nr the circumfer- 
ence of the base of the cylinder. 

Rule. — Multiply the circu???ference of the base by half the 
sum of the greatest and least lengths of the ungula. 

I. What is the convex surface of the cylindric ungula A KB A 
—MM, if AM is 8 feet, BM 12 feet and the radius BE of the 
base 3 feet? 

By formula, S=\{a^a') c lnr=n{a^-a / ) r=n{ 8-f-12) X 3= 
188.49552 sq. ft. 

1. 8 ft.=the least length AN of the ungula, and 

2. 12 ft.=the greatest length BM. 
jj J3. 10 ft.=l(8ft.+12 ft.)=half the sum of the least and 

greatest lengths. 

4. 18.849552 ft.=6^r=the circumferenc of the base. 

5. ,\ 10X18.849552=188.49552 sq. ft.=the convex surface. 



240 FINKEL'S: SOLUTION BOOK. 

III. ' . 188.49552 sq. ft— the convex surface, of the ungula. 

Prob. IiXXXIIJ. To find the volume of a cylindric ungula, 
when the plane passes obliquely through the opposite sides 
of the cylinder. 

Formula.— V==l(a-\-a / )7rr 2 =^7r(a-\-a / )r 2 ,, , 

Rule . — Multiply the Urea of "the base, bf ; half the V least and 
greatest lengths of the ungula. ' . •- 

I. What is the volume of a cylindric ungula whose least 
length is 7 feet, greatest length 11 feet, and- ■' the radius of the 
base 2 feet? ■■• :•.■.■•■-; »fir- --*.?; r» - ;. *'»?.£ ■• • .'• :i - X :'-r 

By formula, F=|(«+« / )7rr 2 =J(7+ll)7r2 2 =113,097312.cu. ft. 

1. 7 ft.=the least length of the ungula, and 

2. irft;===the greatest: length. ■'"',''"' '■' .,','. ,'■... .,;, 

3. 9 ft.=|(7 ft.^-nft;)==half: the length' of tlie^ast f aKd 
IlX greatest lengths. ,' 1( _ lU 

4. 12.566368 sq. ft^===^2 2 =the area of the' basei" ' \ ' 

5. /.9X12.566368=:113.0^7312 cu. ft=the '. volun^otthe 
ungula. 

III. .-. The volume of the ungula is 113.097312 ou. ft; 

Prob. LXXXIV. To find the convex surface of a cylindric 
ungula, when the plane passes through the base and one of 
its sides. ■■ . ; ,-, . t \> >-r-j-:--vi\ :*\i}&zi2 him ;- j --a«: 



\ a r b b—x' ' af. !_*.#, ,-, w, -i*T & 

=2r T / . dx=2r ~ \ b vers l ^-J-^^j^— x 2 '—*' vers - , 

Vo \2rx—x 2 °*- r r Jo 

^^^ir^^^^ir^^Ye^T 1 ^'M%r^l2ri>^b 2 ^\ V 

(^— ^Jvers- 1 -^- 

Rule. — Multiply the sine of half the arc of the base by the 
diatneter of the cylinder, and from the product subtract the prod- 
uct of the arc and cosine; this difference , multiplied by the quo- 
tient of the height divided by the versed sine will be the convex 
surface. 

I. What is the convex surface of the cylindric ungula A CB — 



MENSURATION. 



241 



Z>, whose altitude BD is 28 feet, height BM of arc of base 4 
feet and chord A C 16 feet? 



By formula, S=2rjV ^2rb— b 2 — 
(r— 3)vers- 1 -~1, =2XlOX 

p2Xl0x4— 4 2 — (10— 4)vers- 1 ^"| , = 



28 

4 



140 



:140[8— 5.5638]=341.068 sq. ft. 




FIG. 39. 



11. 



r 1. 28 ft —the altitude BD. 

2. 4 ft.=the height BM of the arc of the base. 

3. 16 ft.=the chord A C of the arc of the base. 

4. 8 ft.===the sine CM oi the arc. 

5. 10 ft.= (8 2 +4 2 )^-(2x4)=the radius OC=OB of the 

base, by Prob. XX, formula R=(a 2 +c 2 )-r-2a. 

6. 6 ft.=10 ft.— 4 ft.=cosine OM of the arc. 

7. 160 sq. ft.=20x8=sine multiplied by the diameter of 
the base. 



ft.=2V8 2 4-4 2 ri4- — — — - ^= 

^ V +60x8 2 +33x4 2 7 



8. 18.5438 ft.=2V8 2 +4 2 ( h& CT»?oow ,=the arc 

60 X 8 - +33 X 

C^^[, by formula of Prob. XXV. 

9. .*. 111.2628 sq. ft.=6Xl8.5438=the arc multiplied by 

the cosine OM. 

10. 160 sq. ft— 111.2628 sq. ft.=48.7372sq.ft.=the differ- 

ence. 

11. .-. 341.1604 sq. ft.=(28H-4)X48.7372sq. ft.=the con- 

vex surface. 



III. .'. The convex surface is 341.1604 sq. ft. nearly. 

Note. — The difference in the two answers is caused by the length of the 
arc CBA, in the solution, only being a near approximation. 

* Demonstration. — In the figure, let BK=x, BM=b, BD=a, 
and the angle BMD=ft. Then MK=b—x, and IK=FL=MK 



tan 0=(b— x)tan#. 



But tanfl„ r/= -. 



, FL=~(b-x). 



Now if we take an elemennt of the arc Z.BU, and from it draw 
a line parallel to FL, we will have an element of the sur- 



242 FINKEL'S SOLUTION BOOK. 

face LBHEGF. This will be a rectangle whose length is 
FLz=~ib — x) and width an element of the arc LBH. An ele- 
ment of the arc is ds=\(dx 2 -\-dy 2 ). Let HK=y. Thenjy 2 = 
2rx — x 2 , by a property of the circle, from which we find dy= 

f $c rdx 

- dx. .'. ds=-—- /. The area of the element of 

\2rx — x 2 \2rx — x 2 



rdx 
'X — 

" b a , , , rdx ~ a T b ,, . dx 



the surface is ~j(b — x) k/ ^ ===> and the whole surface of ABC- 

b x / \2rx — x z 



D is S=2 f \(b-x)-^L=, =24 f (b—x) r 

Jo f S2rx— x 2 bJo ; sl2rx—x 2 

2r a -Xj2rx— x 2 — (r— ^vers" 1 ^"] , ^-Jjlr^rb— b 2 — 
2(r—b)r vers" 1 -"!. Q. E. D. 

Proh, LXXXV. To find the volume of a cylindric ung-ula, 
when the cutting- plane passes througli the hase and one of 
its sides. 

Formula.— V=\ (b—x)dA=j( (b—x)2^2rx—x 2 dx 1 

=2~^(2rx~ x 2 ) % — (r— b) P>J(2rx— x 2 )dx^==2~^(2rx— **)* 

+i(r— x)V2rx— ^2_)_| r 2 sin -i?^Z?+c"| . When %=(?, V=0. 

r -*o 

.-. C=— \nr 2 (r— b). .-. V="T$(2rb— b 2 f— (r— b)\\nr 2 — 
(r—b)V2rb—b 2 —r 2 sin" 1 — ^ I 1. 



Rule. — From §- of the cube of half the chord of the base, sub- 
tract the -product of the area of Hie base and the difference of the 
radius of t/ie base and the height of the arc of the base; this dif- 
ference multiplied by the quotient of the altitude of the ungula by 
the height (versed sine) of the arc of the base, will give the vol- 
ume. 

I. What is the volume of a cylindric ungula, whose altitude 
BD is 8 feet, chord A C of base 6 feet, and height BM oi arc of 
base 1 foot? 



By formula, K=~ \ %(2rb— b 2 f— (r— b) f 



7tr 



MENSURATION. 243 

( r _£)A/(2r3— b 2 ) — r 2 sin 1 ^-~j | =8 J f(2x5Xl— l) f — 
(5— 1) [|7r5 2 — 4^2X5X1=1— 5 2 sin^ 1 ?=!"] | , =8 j 18— 

4["-i7r25— 12— 25 sin^fl j =528+800 sin" 1 ^—200^= 

13.20394 cu. ft. 

'1. 8 ft=the altitude BD. 

2. 1 ft.=the altitude BM of the arc ABC of the base. 

3. 6 ft.=the chord A C of the base. 

4. 18 cu. ft.= f of 3 3 =f of the cube of the sine of half trie 
arc of the base. 

I 3 
II.<!5. 4 T ^ sq. ft.=g — --|-| of 6 Xl=area of the base, by form- 
ula, (b), Prob. XXVIII 

6. 16£ cu. ft.=4Xl T V=the area of the baseX OM, the cosine 
of the arc CHB. 

7. .-. 8(18 cu. ft.— lGi cu. ft.)=13-J cu. ft.=the volume of 
the cylindric ungula ACB — D. 

III. .-. The volume of the cylindric ungula ^4 CB— D is 13^ 
cu. ft., nearly. 

Prob. LXXXVI. To find the convex surface of the frustum 
of a cylindric ungula. 

Formula. — S=i 2rV2r& — b 2 — 2(r — b)r vers -1 - J — 

-^IrVIrT^F 2 — 2(r— b')r vers '- ■ ~|. 

lillle. — (1) Conceive the section to he continued, till it meets 
the side of the cylinder produced; then say, as the difference of 
the heights of the arcs of the two cuds of the ungula, is to the 
height of the arc of the less end, so is the height of the cylinder to 
the part of the side produced. 

(2) Find the surf ace of each of the ungulas, thus formed, 
by Prob. LXXXI \\, and their difference will be the convex sur- 
face of the frustum of the cylindric ungula. 

Prob. LXXXY1I. To find the volume of a frustum of a 
cylindric ungula. 

Formula.— }=^J(2rb— b 2 ) 1 — (r— b) j \nr 2 — 

( r —b)V2rb—b*^r* sin"- 1 ^ I l—^f^rb'— V 2 ) 1 — 



244 



FINKEL'S SOLUTION BOOK. 



{'—!>') \ i7Tr 2 —(r—¥)V(2rb / —b / ' 2 )—r 2 



=?!]■ 



Rule. — Find the volume of the ungula whose base is the the 
upper base of the frustum and altitude that as found by (1) of the 
last rule. Also the volume of the ungula whose base is the lower 
base of the frustum and altitude the sum of the less ungula and 
altitude of the frustum. Their difference will be the volume of 
the frustum. 

3. PYRAMID AND CONE. 

Prob. 1XXXVIII. To find the convex surface of a right 
cone. 



Formula. — S=C s £\h-= c LnrY.\\/ a^-^r' 1 , where C is the 
circumference, h the slant height, r the radius of the base, and a 
the altitude. 

Rule. — Multiply the circumference of the base by the slant 
height and take half the prodw t. Or, if the altitude and radius 
of the base are given, multiply the circumference of the base ly the 
square root of the sum of the squares of the radius and altitude, 
and take half the product. 

I. What is the convex surface of a right cone whose altitude 
is 8 inches and the radius of whose base is 6 inches?. 

By formula, S==2;rrXiV'« 2 +r 2 ==2tf6XiVV+6 2 ==160»= 
188.495559 sq. in. 

1. 6 in.=the radius ^4Z>ofthe base, 
and 

2. 8 in.=the altitude CD. 

3. 10in.=V / 8 2 +6-=the slant 
height CA. 

IU4. 37.6991118 in.=2zrr=l^x 

3.14159265=the circumfer- 
ence of the base. 

4. .-. 188.495559 sq. in= 
i(10x37.6991118)=the con- 
vex surface of the cone. piQ 4 Q 

III. .*. The convex surface of the cone is 188.495559 sq. in. 

Prob. LXXXIX. To find, the convex surface of a pyramid. 

Formula. — S=±p>Xk, in which p is the perimeter of 
the base and h the slant height. 

Rule. — Multiply the perimeter of the base by the slant height 
and take half the product. 




MENSURATION. 245 

I. What is the convex surface of a pentagonal pyramid whose 
slant height is 8 inches and one side of the base 3 inches? 

By formula, S=ip Xh=i( 3+3+3+3+3 ) X 8=60 sq. in. 

'1. 8 in.=the slant height. 

2. 3 in.=the length of one side of the base. 
II. 1 3. 5x3 in. =15 in.=the perimeter of the base. 

4. .'. ^(15X8)=60 sq. in.=the convex surface of the pyra- 
mid. 
III. .*. The convex surface of the pyramid is 60 sq. in. 

Remark. — If the entire surface of a pyramid or cone is required, 
to the convex surface add the area of the base. 

Formula* — T—S-\-A, where A is the area of the base 
and S the convex surface. 

Prob. XC. To find the volume of a pyramid or a cone. 

Formula. — V=$aA=^aX 7r r 2 , where a is the altitude 
and A=7tr* the area of the base. 

Rule. — Multiply the area of the base by the altitude and take 
one-third of the product. 

I. What is the volume of a cone whose altitude CD is 18 
inches and the radius AD ot the base 3 inches? 

By formula, V=\a X7Tr 2 =^X 18 X ?r3 2 =54x 3.14159265= 
169.646 cu. in. 

1. 18 in.= the altitude CD, and 

2. 3 in.=the radius AD. 
II.<j3. 28.274333S5 sq. in.=7rr 2 =3 2 7r=the area of the base. 

4. .-. 169.6460031 cu. in.=^^4=^xl8x3 2 7r=the volume 
of the cone. 
III. .-. The volume of the cone is 169.6460031 cu. in. 

Prob. XCI. To find the convex surface of a frustum of a 
cone. 

F ormula.— S= \{ C+C')k=i(2Trr+2nr')A=: 

7t(r-\-r / )\a 2 -\-(r — r') 2 , in which C is the circumference of the 
lower base, C / the circumference of the upper base, and /*,= 
Va*+(r— r'Y , the slant height. 

Rule. — Multiply half the sum of the circu?nferences of the 
two bases by the slant height. 

1. What is the convex surface of the frustum of a cone whose 
altitude is 4 feet, radius of the lower base 4 feet, and the radius of 
the upper base 1 foot? 



246 FINKEL'S SOLUTION BOOK. 



By formula, 5=^(r+r / )V« 8 + (r— r / )2=^(4+l)V4 2 + (4— 1) ; 
=25^=78.539816 sq. ft. 

1. 4 ft.= the altitude OB, 

2. 4 ft.=the radius AE of the lower 
base, and 

3. 1 ft.=the radius DO of the upper 

4. 3 ft.= AB— PE(=D 0)= r —r'. 

5. 5 ft.=<J(jDP*+AJP*)= 



II. 



A e^ 



V^2 + ( r _^) 2;= V4 2 +(4— l) 2 
. =AB>, the slant height. 

6. 87T=the circumference AGBHoi 
the lower base. ^ /G 4 '' 

7. 27f=the circumference DIC of the upper base. 

8. 5^=-J(8T-(-27r )=half the sum of the circumferences. 

9. .*. 5 X57r=257T=78. 539816 sq. ft.=the convex surface of 
the frustum. 

III. .*. The convex surface of the frustum is 78.539816 sq. ft. 

Remark. — If the entire surface of the frustum is required, to 
the convex surface add the area of the two bases. 

Formula.— T=S+A+A / =7r(rJ r r / )\laZ + (r— r'y + 
7tr 2 -\-7tr /2 . 

Prob. XCII. To find the convex surface of the frustum of 
a pyramid. 

Formula.— S=\{p+f )h. 

Rule. — Multiply half the sum of the perimeters of the two 
bases by the slant height. 

I. What is the convex surface of the frustum of a pentagonal 
pyramid, if each side of the lower base is 5 feet, each side of the 
upper base 1 foot, and the altitude of the frustum 10 feet? 

Before we can apply the formula, we must find the slant 
height. Produce BO, till 0K=0E. Divide OK into extreme 
and mean ratio at H. Draw BH. Then KO \OH::OH :KH. 
.-. OH*=KOxKH=KOx{K0—OH)=K0 2 —KOxOH; > 
whence OJ7 2 -\-KOx OH=K0 2 . Completing the square of this 
equation, OH 2 -\-KOxOH+\K0 2 =\KO* , from which OH( = 
EH=BK)=%KO(slb—l). EB 2 =EK 2 —KB*=\\KO(4h— 

\)^—\\{KO— OH)Y,-=\ KO *(4l— l) 2 -[i ) KO—±KO(sll— 
1)| ] =±K0 2 (*JE— l) 2 — i^0 2 (3— V5) 2 =i^T0 2 £(V5— l) 2 — 



MENSURATION. 247 

i(Z-jJ5)*^=iirO* [^12l^-2^|= T i^02 (10—2^6). But FF= 
$EA=is. .'. is*=£rKO 2 (10— 2V / 5), and fc^/TO^lO— 2Vd. 

.*. KO= , where j is a side of the lower base,= 

V10— 2^5 



KG may be considered the radius R of a circui 



VI0-2V5 

scribed circle of the lower base. In like manner, the radius r of 

the circumscribed circle of the upper base may be found to be 

2s' 2 

— — where s' is a side of the upper base,= 

VlO— 2Vg V10—2VI 

OF, the apothem of the lower base, =V (F 2 —FF 2 )= 

+ [ 2 n( j^ 1 — HYV ( =fV650+W5= the slant height. 

By formula, S=£( 25+5) 1^650+10^5=6^650+10^= 

155.5795 sq. ft. 

fl. 10 ft.=the altitude <?£. 

2. 5 ft.=Z?-<4, one of the equal sides of 
the lower base. 

3. 1 ft.=ed, one of the equal sides of 
the upper base. 



II. 




4. fV650+l(V5=/^\ the slant height. 

5. 5X5 ft.=25 ft.=the perimeter of the 
lower base. 

6. 5X1 ft.=5 ft— the perimeter of the 
upper base. r ^' 

7. .-. i(25+5)|V650+10V5== 155.5795 sq.ft.=the convex 
surface. 

III. .*. The convex surface of the frustum is 155.5795 sq. ft. 

Prob. XCIII. To find the volume of a frustum of a pyra- 
mid or a cone. 



II. 



248 FINKEL'S SOLUTION BOOK. 

Formula.— (a) V=±a(A+V 'AA'-\-A') f in which A 
is the area of the lower base, A / the area of the upper base and 
V AA / the area of the mean base. When we have a frustum of 
acone, (b) V=,^a(A-\-^A^ / +A / )^=ia(7tR 2 -\-V( ttR 2 X nr 2 )+ 
irr 2 )=ia(7t R 2 -\-7t Rr-\- 7tr 2 )=^7t a(R 2 +Rr-\-r 2 ). 

Rule. — (1) Find the area of the mean bast by multiplying the 
area of the upper and lower bases together and extracting the 
square root of the product. 

(2) Add the upper, lower, and mean bases together and multi- 
ply the sum by -J the altitude. 

I. What is the solidity of a frustum of a cone whose altitude 
is 8 feet, the rad:us of the lower base 2 feet, and the radius of 
the upper base 1 foot? 

By formula (b), V=i 7 ra(R' 2 +Rr-\-r ,i )=i 7 r 8(4+2+1)= 
■J X 56 ?r =58.6433 cu. ft. 

1. 8 ft.=the altitude. 

2. 2 ft.=the radius of the lower base. 

3. 1 ft.=the radius of the upper base. 

4. 4 7T=the area of the lower base. 

5. 7T=the area of the upper base. 

6. 27t=^4:7t X 7r=the area of the mean base. 

7. 4 7T+ 7T+2 7t=l7f =the sum of the areas of the three bases. 

8. /. iXSx77t =58.6433 cu. ft.=the solidity of the frustum. 
III. .-. Ths solidity of the frustum is 58.6433 cu. ft 

4. CONICAL UNGULAS. 

1. A Conical Ungula (Latungula, a claw, hoof, from 
unguis, a nail, claw, hoof) is a section or part of a cone cut off 
by a plane oblique to the base and contained between this plane 
and the base. 

Prob. XCIV. To find the surface of a conical ung-ula. 

/•* , 

Formula. — SI s\dx 2 -\-dy 2 = 

_, r(2R—t)r—(R+r—t)x-\ ) _ . . . , . , 

2* cos M v ! — I } dx, where a is the altitude 

of the ungula, R the radius of the base, r the radius of the upper 
base of the frustum from which the ungula is cut, t the distance 
the catting plane cuts the base from the opposite extremity of 
the base, and x the radius of a section parallel to the base and at 
a distance h — y from the base. 



MENSURATION. 



24$ 



Prob. XC V. To find the volume of a conical ungula. 
Formula. — V— I Ady= 
■J r ( (R — r)x (R — r) 2 L v 



(R — r)x 
_(7?_|_ r _/) A n^ |Q_(2i?_ ty r 2 +2r(2R— t)(R-\-r— t)x— 

(2R—t)(2r— t)x 2 l | dx, 

where the letters represent the same value as in the preceding 

Li i7 / « \, • a(x — r) 
problem and dy=l -=, — ; \dx, since y=~~ -. 

Prob. XCVI. To find the convex surface of a conical un- 
gula, when the cutting" plane passes through the opposite 
extremities of the ends of the frustum. 

Formula. — S== 

■j£z?Va*+{R-r)* j R*-i(R+r)VR? \ . 

This formula is obtained by putting / ==#, in 
the formula of Prob. XCIV., and integrating the 
result. For, in this problem, the cutting plane 
AHCK passes through the opposite point A, and 
therefore the distance from A to the cutting 
plane is 0. .-. /==0. 

FIG. 43. 
Rule. — Multiply half the sum of the radii of the bases by the 
square root of their prod net and subtract the result from the 
square of the radius of the lower base. Multiply this difference 
by n times the slant height and divide the result thus obtained by 
the difference of the radii of the bases. 

Prob. XCVI I. To find the volume of a conical ungula, 
when the cutting plane passes through the opposite extremi- 
ties of the ends of the frustum. 




Formula. — I = 



:>( 



nfiiSa s 3 3 \ 



This formula is obtained by putting & 
XCV., and integrating the result. 



-0, in the formula of Prob. 



Rule. — Multiply the difference of the square roots of the 
cubes of the radii of the bases by the square root of the cube of the 
radius of the lower base and this product by -J-tt times the altitude. 



250 FINKEL'S SOLUTION BOOK. 

Divide this last product by the difference of the radii of the two 
bases and the quotient will be the volume of the ungula. 

I. A cup in the form of a frustum of a cone is 7 in. in diame- 
ter at the top, 4 in. at the bottom, and 6 in. deep. If, when full 
of water, it is tipped just so that the raised edge of the bottom 
is visible; what is the volume of the water poured out? 

By formula, V=^^(r'— r f )=J^(49— 8^7)= 

102.016989 cu. in. 

Re?nark. — Fig. 43 inverted represents the form of the cup and 
APBQ — C the quantity of water poured out, C being the tipped 
edge of the bottom. 

I. A tank is 6 feet in diameter at the top, 8 feet at the bot- 
tom, and 12 feet deep. A plane passes from the top on one side 
to the bottom on the other side : into what segments does it 
divide the tank? 

B> fo,-n,,la, ,-=^(^-,1)^^(8-3^),= 
32?r(8-3V3)=281.87 cti ft. 



lU 



1. 4 ft.— AH, the radius of the lower base. 

2. 3 ft.=Z>/ 7 , the radius of the upper base, and 

3. 12 ft=jFZ, the altitude. Then 

4. W ^ X ^ (V^-V^)=327r(S-W^)=2Sl.S7 



sq. ft.=the volume. 
III. .'. The volume is 281.87 cu. ft. 

Prob. XCVIII. To find the convex surface ofa conical un- 
gula, when the cutting- plane FCE makes an angle CIB less 
than the angle DAB, i. e. when AI(=t) is less than DC(==2r). 

FormiUa.—S=j^-^ 2 +(J?—r) 2 j ^ 2cos " 1 ( : § : - / ) 

2r—t y } v ' 2r—t ^ 2r—t \ r J 

This formula is obtained by integrating the formula of 
Prob. XCIV, recollecting that the co- efficient of x 2 is negative. 

Prob. XCIX. To find the volume of a conical ungula, when 
the cutting 1 plane FCE makes an angle CIB less than the an- 
gle DAB. i. e., when AI (=t) is less than CD (=2r). 



MENSURATION. 



251 



R+t\ 



Formula* — V^=jr^T ) «^ 3cos | ' — ) 



2R—t)t- 



(R+r-t)(R-r) 

X 



», 



n /?r(i?— r )-| ij 

:i L t— 2r J^ — '/' ' (2R—t)(t—-2r) 

[(^)']'^ e ( "^y r "" ) 

This formula is obtained by integrating the 
formula of Prob. XCV, recollecting that the co- FIG. 44. 

efficient of at 2 is negative. 

Prob. C. To find the convex surface of a conical un- 
gula, when the cutting plane FCE is parallel to the side AD, 
i. e., when AI(=t) is equal to DC(=2r). 

Farmtaa.—S==^-^Ja^+( /?—/-)■-* \ R- c ^~ 1 (~^- 2 -) 

+-2(R— 2r)\ , (R— r)r— %{R— r)\ \R-^r)- ' 



°£^ 



This formula is obtained by putting /=2r, 
in the formula of Prob. XCIV., and inte- 
grating the resulting equation. 

Pol). CI. To find the volume of a coni- 
cal ungmla, when the cutting planeFCE 
is parallel to the side DA, i„ < ., when 
AI(=t) is equal to CD (==2r). 

Formula.- V=$a j -^^^^0^^^+ 

This formula is obtained by putting t=2r, in the formula of 
Prob. XCV., and integrating the resulting equation. 

Proh. CII. To find f he convex surface of a concial ungula, 
when the cutting plane FCE makes an angle C1B greater 
than the angle DAK, i. e., when AI (=*) is greater than DC 



: '0^ B 



FIG 45. 



Formula.— S=. 






252 FINKEL'S SOLUTION BOOK. 

This formula is obtained by integrating the formula of Prob. 
XCIV., remembering that the coefficient of x 2 , which occurs in 
process of integrating, is positive. 

Prob. CIII. To find the volume of a conical ungula, when 
the cutting- plane FCE makes an angle CIB greater than the 
angle DAB, i. e., when AI(=t) is greater than DC(=2r). 

Formula.— V==j^zp j i^ 8 cos-*(-^t')— 

* r \ (t—2ry J^t—2r X 

lo g [(,_, +( ,_ 2 ,)J_l_^]. 



°m^ c 



FIG. 46. 



This formula is obtained by integrating the formula of Prob. 
XCV., regarding the coefficient of* 2 positive. 

XII. THE SPHERE. 
Prob. CIV. To find the convex surface of a sphere. 

Formula.— S=2 Y s 2ny^dy 2 ^dx^=^7tR' l ^= ttZ> 2 , where 
D is the diameter. 

Rule. — Multiply the square of the diameter by 3.1^.1592. 
I. What is the surface of a sphere whose radius is 5 inches? 

By formula, 6 , =4^i? 2 =47rX 25=314.1592 sq. in. 
"1. 5 in.=the radius. 

2. 25 sq. in.= the square of the radius. 

3. .'. 4^X25 sq. in.=314.1592 sq. in.=the surface of the 
sphere. 

III. .*. 314.1592 sq. in. =the surface of the sphere. 

Note. — Since 7i\ff 2 is the area of a circle whose radius is i?, the area 
(4?ri? 2 ) of a sphere is equal to four great circles of the sphere. The sur- 
face of a sphere is also equal to the convex* surface of its circumscribing 
cylinder. 

Prob. CV. To find the volume of a sphere, or a globe. 
Formula.— V=2ity*dx=$n R*=%it{\D)*=%itD*. 

Rnle. — Multiply the cube of the radius by %n (=^.188782); or 

multiply the cube of the diameter by \r (=.5285987). 

L What is the volume of a sphere whose diameter is 4 feet? 



II. 



II 



MENSURATION. 253 

By formula, F=|7ri? 3 =|^2 3 =33.510256 cu. ft. 

1. 2 ft.=the radius. 

2. 8 cu. ft.=2 3 =the cube o f the radius. 

3. .-. 4.188782x8 cu. ft.=33.510256 cu. ft.=the volume of 
the sphere. 

III. .-. 33.510256 cu. ft.=the volume of the sphere. 

Prob. CVI. To find the area of a zone. 

A. ZoTie is the curved surface of a sphere included between two 
parallel planes or cut off by one plane. 

Formula. — S—%7tRa, in which a is the altitude of the 
segment of which the zone is the curved surface. 

Rule. — Multiply the circi*mfercnce of a great circle of the 
sphere by the altitude of the segment. 

I. What is the area of a zone whose altitude is 2 feet, on a 
sphere whose radius is 6 feet? 

By formula, S=2nRa=2 ttBx 2=24 n =76.39822 sq. ft. 

1. 6 ft.=the radius of the sphere. 

2. 2 ft.=the altitude. 
II. s 3. 12 7r=37. 69911 ft.=the circumference of a great circle of 

the sphere. 

4. .-.2X3769911=75.39822 sq. ft.=the area of the zone. 
III. .-. The area of the zone is 75.39822 sq. ft. 

Note. — This rule is applicable whether the zone is the curved surface of 
the frustum of a sphere or the curved surface of a segment of a sphere. 

Prob. CVTI. To find the volume of the segment of a sphere. 

Formula, — V=±7ra(or\-\-a' 2 ) where r 1 is the radius or 
the base of the segment. 

Rill©. — To three li?ues the square of the radius of the base, add 
the square of the altitude and multiply the sunt by ^n=.52359S7 
ti?nes the altitude. 

I. What is the volume of a segment whose altitude is 2 inches 
and the radius of the base 8 inches? 

By formula, Fz=-J^(3;-;+^ 2 )=^X2(3X64+4)=205.2406 
cu. in. 

1. 8 in.=the radius of the base. 

2. 2 in.=the altitude of the segment. 

3. 192 sq. in = i 3X^"= :: three limes the square of the radius. 
II. i 4. 4 sq. in.=the square of the altitude. 

5. 196 sq. in. =192 sq. in.-|-4 sq. in. =three times the square 
of the radius plus the square of the altitude. 

6. \n X-XK)6=205. 2406 cu. in.=the volume of the segment- 



254 FINKEL'S SOLUTION BOOK. 

III. .*. 205.2406 cu. in ==the volume of the segment. 

Note. — From the formula V r =^ / ira(Sr lL -f-d z ), -we have V^lTrarj^-^^wa 3 
But \xar\ is the volume of a cylinder whose radius is7-j, and altitude \a, and 
^7r«3 i s the volume of a sphere whose diameter is a :. The volume of a 
segment of a sphere is equal to a cylinder whose base is the base of the seg- 
ment and altitune half the altitude of the segment, plus a sphere whose 
diameter is the altitude of the segment. 

Prob. CVIII. To find the volume of a frustum of a 
sphere ,or the portion included between two parallel planes. 

Formula. — V=\7ia\Z(?']A r ?-\)-) r a 2 \=±a(7ir\-\-7irl)-\- 
^7ta z *, in which ry is the radius of the lower base, r 2 the ra- 
dius of. the upper base. 

Rule. — To three times the sum of the squared radii of the two 
ends, add the square of the altitude; multiply this sum by .5235987 
times the altitude. 

I. What is the volume of the frustum of a sphere, the radius of 
whose upper base is 2 feet and lower base 3 feet a ud altitude \ 
foot? 

By formula, V=\ *43(rf+r5)+fl»]~=4 , .X+[3(9+4)+±]= 
8.03839 cu. ft. 

1. 3 ft.=the radius of the lower base. 

2. 2 ft.=the radius of the upper ba>e. 

3. 39 sq. ft.=:3(3 2 +2 2 )=three times the sum of the squares 
II. <| of the radii of the two bases. ; 

4. \ sq. ft.=the square of the altitude. 

5. .-. -i-Tr X^X39i=8.03839 cu. ft.=the volume of the 
fruMum. 

III. .'. 8.03839 cu. ft —the volume of the frustum. 

Prob. CIX. To find the volume of spherical sector. 

A. Spherical Sector is the volume generated by any sector 
of a semi-circle which is revolved about ils diameter. 

Formula. — V =\it aR' 1 , where a is the altitude of the 
zone of the sector. 

Rtlle. — Multiply its zone by one-third the radius. 

*Note. — -!tf(7r/-j-|-7r7' 2 )— the volume of two cylinders whose 'bases' are 
the upper and lower bases of the segment and whos. 1 altitude is luili the alti- 
tude ot the segment. \~u z is the volume of a sphere whose diameter is the 
altitude ot the segment. Hence the volume of a segment of a sphere of two 
bases H equivalent to the volume of two cylinders whose bases are the up- 
per and lower bases respectively of the segment and whose common altitude 
is the altitude of the segment, plus the volume of a sphere whose diameter 
is the altitude of the segment. 

For a demonstration of this and the preceding formula, s.e WentivortKs 
Plane and Solid Geometry, Bk. IX., Prob. XXXII. 



MENSURATION. 



255 



I. What is the volume of a spherical sector the altitude of 
whose zone is 2 meters and the radius of the sphere 6 meters? 

By formula, V=\naR*=\x X2x6 2 = 
150.7964m 3 . 

1. 2m.=the altitude BD of the zone gener- 
ated by the arc EF when the semicir- 
cle is revolved about AB. 

2. 6m.=the radius EC of the sphere. 

3. 2 7r6m.=37. 699104 m =the circumference 
of a great circle of the sphere. 

4. 2^-6X2=75.398208 m 2 .=the area of the 
zone generated by EE, by Prob. CVI. 

5. .-. iX6x75.39S208=150.796416nvWthe volume of the 
spherical sector. 

.-. The volume of the spherical sector is 150.796416 m 3 . 



11^ 




FIG. 47. 



Ill 



I. Find the diameter of a sphere of which a sector contains 
7853.98 cu. ft., when the altitude of its zone is 6 feet. 

By formula, V=%7tar 2 =%7t X6X^ 2 . .*. f arx6X^ 8 = 

7853.98 cu. ft, or 4r 2 =2500 sq. ft., whence 2r=b0 feet, the 
diameter of the sphere. 

'1. 6 ft.=the altitude of the zone. 

2. .*. \ir x6X ? ' 2 =the volume of the sector. But 

3. 7853.98 cu. ft.=the volume. 

4. .-. |7rX6X^ 2 =7853.98 cu. ft. 

5. r 2 =625 sq. ft, by dividing by An. 
Q. .'. 2r=50 ft., the diameter of the sphere. 

III. .-. The diameter of the sphere is 50 feet. 

Prob. CX. To find the area of a lime. 

A Lane is that portion of a sphere comprised between two 
great semi-circles. 



II. 



Formula. — 5=4 rr R 2 l ( — d )==47rR 2 u, where 



u is 



J 

the quotient of the angle of the lune divided by 360°. 

Rule. — Multiply the surface of the sphere by the quotient oj 
the angle of the lune divided by 360° 

I. Given the radius of a sphere 10 inches; find the area of a 
lune whose angle is 30°. 

By formula, S=±nR* n=A X * XlO 2 X(30°-h360° )— 
J?rl0 2 =104.7197 sq. in. 



II 



256 FINKEL'S SOLUTION BOOK 

1. 10 in.=the radius of the sphere. 

2. 30°=the angle of the lune. 

3. T i2=30°-r-360°=-the quotient of the angle of the lune 
divided by 360°. 

4. 4 7rl0 2 =400 7T=1256.6368 sq. in.=the suiface of the 
sphere. 

5. .-. T VX 1256.6368 sq. in.=104.7198 sq. in.==the area of the 
lune. 

III. .-. The area of the lune is 104.7198 sq. in. 

Wentworth 's New Plane and Solid Geometry , p. 371, Ex. 585. 

Prob^ CXI. To find the volume of a spherical ung-ula. 

A Spherical JJngula is a portion of a sphere bunded by a 
lune and two great semi-circles. 

Formula. — V=^7t R z u, where u is the same as in the 
last problem. 

Rllle. — Multiply the area of the lune by one-third the radius ; 
or, multiply the volume of the sphere by the quotient of the angle 
of the lune divided by 360°. 

I. What is the volume of a spherical ungula the angle of 
whose lune is 20°, if the radius of the sphere is 3 feet? 

By formula, V=.\n R* u=\n X 3 3 X (20°-f-360°) = 6.283184 
cu. ft. 

1. 3 ft.=the radius of the sphere. 

2. 47r3 2 X(20°-^360°)=6.283184 sq. ft.=the area of the 
II. <J lune, by Prob CX 

3. .*. iX3x6.283184=6.283184 cu. ft.=the volume of the 
ungula. 

III. .-. 6.283184 cu. ft. is the volume of the ungula. 

Prob. CXII. To find the area of a spherical triangle. 

Formula.— S=2?t R* x (A+B+C— 180°)-r-360°, in 
which A, B, and C are the angles of the spherical triangle. 

K-llle. — Mtdtiply the area of the hemisphere in which the tri- 
angle is situated by the quotient oj the spherical excess (the ex- 
cess of the sum of the spherical angles over 180°) divided by 360°. 

I. What is the area of a spherical triangle on a sphere 
whose diameter is 12, the angles of the triangle being 82°, 98°, 
and 100°? 

By formula, S=2 ttR 2 x(A+B+C— lS0°)-r-360°=27r 6 2 X 
(82°+98°+100°— 180 o ;-^360 o =27r6 2 XfV=62.83184=area. 



II. 



MENSURATION. 257 

'1. 6=the radius of the sphere. 

2. 2ft 6 2 =72 7r=the area of the hemisphere. 

3. (82°-f-98°+100°— 180°)=100°=the spherical excess. 

4. 100 °-r-360°= T 5 8=the quotient of the spherical excess 
divided by 360°. 

.5. .*. T \X727r=62.83184=the area of the spherical triangle. 
III. .". The area of the spherical triangle is 62.83184. 
(Olney's Geometry and Trigonometry, Un. Bd.,p.238,Bx. 8.) 
Prob. CXIII. To find the volume of a spherical pyramid. 

A Spherical Pyramid is the portion of a sphere bounded 
by a spherical polygon and the planes of its sides. 

Formula.— V=*7rR*X(&^-360 o ), where B is the 
spherical excess. 

Rule. — 'Multiply the area of the base by one-third of the radius 
of the sphere 

I. The angles of a triangle, on a sphere whose radius is 9 feet, 
are 100°, 115°, and 120° ; find the area of the triangle and 
the volume of the corresponding spherical pyramid. 

By formula, V=,\n R* x (i±-f-360° )=f*i? 3 X {A+B+C— 
180 o )-f-360° = |7r9' 3 X (100°+115 o +126°— 180°)-r-360° = 
^«r9 3 =657.377126 cu. ft. 

1. 9 ft.=the radius of the sphere. 

2. 2?r9 2 — the area of the hemisphere in which the pyramid 
is situated. 

3 (100°+115 o +120°— 180°)=155°=the sperical ex- 
cess. 
II. < 4. &J— 155 °-i-36G =the quotient of the spherical excess 
divided by 360°. 

5. .'. f£X27r9 2 =f^X ar 9 2 =the area of the base of the pyra- 
mid. 

6. .-. £x9x!iX27r9 2 =657.377126 cu. ft.=the volume of 
the pyramid. 

III. .-. The volume of the spherical pyramid is 657.377126 
cu. ft. 

( Van Amringe's Davies' Geometry and Trigonometry, p. 278, 
Ex. 15. 

I. Fmd the area of a spherical hexagon whose angles are 96 ° , 
110°, 128°, 136°, 140°, and 150°, if the circumference of a 
great circle of the sphere is 10 inches. 

xr ico n*l T —( n — 2)180°] . _, . 

Formula. — S=27rfi 2 ± v 7 J , where T is 

the sum of the angles of the polygon and n the number of 
sides. 



258 FINKEL'S SOLUTION BOOK. 

By for^la, 5 = 2 ^ X f^y^l=2,x(g) 2 X 

(96° +110° +128° +136° +140° +150°— (6— 2)xl80°)-r- 

50 50 

360°== — X(760°— 720° )-v-360°=|— =1.7684 sq. in. 

1. 5-i-7r=the radius of the sphere, since 2ttR=10 in.. 

2. 760°=96 o +110 o +128 o +136 o +140 o +150°=the 
sum of the angles of the polygon. 

3. 760°— (6— 2)xl80°=40°=the spherical excess. 

4. ^=40°-r-360°=the quotient of the spherical excess di- 
vided by 360°. 

(5 A 2 
— 1 =the area of the hemisphere on which the 

polygon is situated. 
6. .-. \Y.2n( - V=4x50-f-jr=1.7684 sq. in. 

III. .-. The area of the polygon is 1.7684 sq. in. 
WentwortK s Geometry, Revised Ed.,f. 374, Ex. 596. 



II. 



XIII. SPHENOID. 

1. A. Spheroid, is a so]id formed by revolving an ellipse 
about one of its diameters as an axis of revolution. 

1. THE PROLATE SPHEROID. 

1. The Prolate Spheroid is the spheroid formed by in- 
volving an ellipse about its transverse diameter as an axis <<f 
revolution. 

Prob. CXIV. To find the surface of a prolate spheroid 

I dy 2 
Formulae. — (a)S=2 C27iy-ds=2 C2 ny ^i_u^- g dx= 

47r- (a 2 — e 2 x 2 ) dx=2nb 2 -\-2 sin' 1 e,=27rb\ ' b-\ — sin -1 *?), where 

ct e e 

Va 2 b 2 

e= —the eccentricitv of the ellipse which generates the 

a 

surface. 

{b ) S^ 7 rab(l---—-^^-^ IIM -&c.) 



MENSURATION. 259 

Rule. — Multiply the circumference of a circle whose radius 
is the semi-conjugate diajncter by the semi-conjugate diameter in- 
creased by the product of the arc whose sine is the eccentricity into 
the quotient of the semi-transverse diameter divided by the eccen- 
tricity. 

I. Find the surface of a prolate spheroid whose transverse 
diameter is 10 feet and conjugate diameter 8 feet. 

By formula (a), S=27rb(b-{-~ si n - 1 c)=27r4(4+- sin-^)= 

Cs V5 2 — 4 2 x y'52 — 42-1 

4 +C 5 ^-^5— ) sin_1 -^ T ^J=2 7 r4[4+ 

(5~|)sin- 1 l]=|7r[48+100sin- 1 f]==|^[48+100X^^^]== 

|tt[48+100x. 6485053] =235.3064 sq. ft. 

1. 25.1327412=27r4=the circumference of a circle whose 
radium is the semi -conjugate diameter of the ellipse. 

V5 2 4- 

2. 4= =the eccentricity. 

5 J 

3. 2 ^ 5 ft. =5 ft-f-f=the quotient of the semi-transverse diame- 
ter divided by the eccentricity. 

II. J 4. .6435053=the arc(to the radius 1 ) whose sine is f , or the 
eccentricity. 

5. 5.3625442 ft.=yft. X. 6435053= V ft. X the arc whose 
sine is | 

6. 9.3625442 ft.=4 ft.+5.3625442 ft.=semi - conjugate di- 
ameter increased by said product. 

7. .-. 235.306.4 sq. ft.=9.3625442X25.1327412=the surface 
of the prolate spheroid. 

1.11. .'. The surface of the prolate spheroid is 235.3064 sq. ft. 

Prol>. CXV. To find the volume of a prolate spheroid. 

b 2 f" 
Formula. — V= f7ry 2 dx=7r — I (a 2 — x 2 )dx— 
<J a J -a 

&* r ~i" 

^—7,1 <*"* — i** =%7rb 2 aj in which b is the semi-conjugate 

a ~ I— _J -it 



diameter, and a the semi-transverse diameter. 

Rule. — Multiply the square of the sc/ui-con/ugate diameter by 
the semi-transverse- diameter and this product by \ri. 

I. What is the volume of a prolate gpheroid, whose semi-trans- 
verse diameter is 50 inches, and semi-conjugate diameter 30 inches. 

By formula, V=\nb 2 a=\n%W X 50=188495.559 cu. in. 



260 FINKEL'S SOLUTION BOOK. 

1. 30 in.=the semi -conjugate diameter, 

2. 50 in.=the semi- transverse diameter. 

3. 900 sq. in— the square of the semi-conjugate diameter. 
II.<! 4. 45000 cu. in.=50x900=the squareofche semi-conjugate 

diameter by the semi-transverse diameter. 
5. .•.|jr45000=|X3.14159265x45000cu.in.= 

188495.559 cu. in.— the volume of the prolate spheroid. 
III. .•. The volume of the prolate spheroid is 188495.559 cu. in. 

2. THE OBLATE SPHEROID. 

1. An Oblate Spheroid is the spheroid formed by revolving 
an ellipse about its conjugate diameter as an axis of revolution. 

Prob. CXVX. To find the surfae of an oblate spheroid. 



dx 2 
Formulae. — (a) S= C27txds=2 I 27rx^l-\- — dy= 



,=2 f a 27rx^\: 

J -a 



e 2 e* 3<? 6 3 be s 

Prob. CXVII. To find the volume of an oblate spheroid. 

Formula.— V=j7tx 2 dy=2f 7t~ 2 (b 2 —y 2 )dy=±7ta 2 b. 

Rule. — Multiply the square of the semi-transverse dia?neter 
by the semi.conjugae diameter and this product by ^n. 

I. What is the volume of an oblate spheroid, whose trans- 
verse diameter is 100 and conjugate diameter 60? 

By formula, V=±na?b=% tt50 2 X30=314159.265. 
'1. 30=^- of 60= the semi-conjugate diameter. 

2. 50=-^ of 100=the semi-transverse diameter. 

3. 2500=50 2 =the square of the semi-transverse diameter. 
II.<[4. 75000=30 X2500=the square of the semi-transverse di- 
ameter multiplied by the semi-conjugate diameter. 

5. .'. |*rX 75000=314159.265= the volume of the oblate 
spheroid. 
III. .-. The volume of the oblate spheroid is 314159.265. 

Note. — Since the volume of a prolate spheroid is \-Kb % a. We may write 
%Trb t a=\{irb' t X.2(i). But Trb 2 X2a is the volume of a cylinder the radius of 
whose base is b and altitude 2a. .'. The volume of a prolate spheroid is § of 
the circumscribed cylinder. In like manner, it may be shown that the vol- 
ume of an oblate spheroid is f of its circumscribed cylinder. 

• 

The following is a general rule for finding the volume of a 
spheroid; Multiply the square of the revolving axis by the fxed 
axis and this product by \it. 



MENSURATION. 



261 



Prob. CXVIII. To find the volume of the middle frustum 
of a prolate spheroid, its length, the middle diameter, and 
that of either of the ends being- given. 

CASE I. 

When the ends are circular, or parallel to the revolving axis. 

JPoV7Yl r ula — V=-Yzir(2jD 2 -\-d 2 )l, where D is the middle 
diameter CD, d the diameter HI of an end, and / the length of 
the frustum. 

Rule. — To twice the square of the middle diameter add the 
square of the diameter of either end and this sum multiplied by 
the length of thefrustu??i y and the product again by y 1 ^ *, will 
give the solidity. 

I. What is the volume of the middle frustum HIGFoin. 
prolate spheroid, if the middle diameter CD is 50 inches, and 
that of either of the ends HI or FG is 40 
OK IS inches? 



inches, and its length 




By formula, V=^n (2Z> 2 +^)/= T ^ (2 X50 2 +40 2 )18= 
31101.767265 cu. in. 

1. 50 in.=the middle diame- 
ter CD. 

2. 40 in.=the diameter of ei- 
ther end as HI 

3. 18 in.=the length OK of 
the frustum. 

4. 5000sq.in.=2x50 2 = twice 
II. < the square of the middle 

diameter. 

5. 1600sq.in.=40 2 =the 
sqaureof the diameter of either end. 

6. 5000 sq. in. +1600 sq. in.=6600 sq. in. 

7. 18X6600=118800 cu. in. 

8. .-. T V n XI 18800 cu. in.=31101.767265 cu. in— the vol- 
ume. 

III. .-. The volume of the frustum is 31101.767265 cu. in. 

CASE II. 

When the ends are elliptical, or perpendicular to the revolving 
axis. 

Formula.— V=^n(2Dd+D / d / )l i where D and d are 
the transverse and conjugate diameters of the middle section and 
D / and d' the transverse and conjugate diameter of the ends and 
/ the distance between the ends. 

Rule. — ( 1 ) Multiply twice the transverse diameter of the 
middle section by its conjugate diatnetcr y and to this product add 



II. 



262 FINKEL'S SOLUTION BOOK. 

the -product of the transverse and conjugate diameter of either of 
the ends. 

(2) Multiply the su?n, thus found, by the distance of the ends, 
or the height of the frustum, and the product again by ^n and 
the result will be the volume. 

I. What is the volume of the middle frustum of an oblate 
spheroid, the diameter of the middle section being 100 inches 
and 60 inches; those of the end 60 inches and 36 inches ; and the 
length 80 inches? 

By formula, V= T 1 J 7r(2I)d-{-I) / d / ) /= T Vr(2x 100x60+60 X 
36) 80=296566.44616 cu. in. 

1. 100 in.=the transverse diameter FC of the middle section. 

2. 60 in. =the conjugate diameter ms of the middle section. 

3. 12000 sq. in.=2X 100 X60=twice the product of the 
diameters of the middle section. 

4. 60 in.=the transverse diameter 
AB of the end. 

5. 36 in.=the conjugate diameter 
2(nc) of the end. 

6. 2160 sq. in.=the product of the 
diameters of the end. 

7. 14160 sq. in.=12000 sq. in.+2160 
sq. in. 

8. 80X14160=1132800 cu. in. = the FIG. 49. 
product of said sum by the height of the frustum. 

9. .-. T V*X 1132800 cu. in.=296566.44616 cu. in.=the vol- 
ume of the frustum. 

III. .-. The volume of the frustum is 296566.44616 cu. in. 

Prob. CXIX. To find the volume of a segment of a prolate 
spheroid 

CASE I. 

When the base is parallel to the revolving axis. 

Formula.— V=\7th 2 (y\ (3Z>— 2/*), where h is the 

height of the segment, d the revolving axis, and D the fixed 
axis. 

Rule. — (1) Divide the square of the revolving axis by the 
square of the fixed axis, and multiply the quotient by the differ- 
ence between three times the fixed axis and twice the heignt of 
the segment. 

(2) Multiply the product, thus found, by the square of the 
height of the seg?nent, and this product by \n, and the result will 
be the volume of the segment. 




MENSURATION. 263 

I. What is the volume of a segment of a prolate spheroid of 
which the fixed axis is 10 feet and the revolving axis 6 feet and 
the height of the segment 1 foot? 

By formula, V=**h*f^\* ($£>•— 2h)= 

^ 7rx62 (S) 2(3xio ~ 2xi)= 

5.277875652 cu. ft. A 

1. 10 ft.=the transverse diameter f 
2BF. D EfZ 

2. 6 ft.=the conjugate diameter AE. /^-^_;— — -"\ 

3. ^9- 5 - = =— — =the square of the conju- 
gate diameter divided by the FIG. 50 
square of the transverse diameter. 

4. 28ft.=3XlO ft.— 2X1 ft— the difference between three 
times the transverse diameter and twice the height of 
the segment. 

5. / T X28 ft.=10&.ft.=the product of said quotient by 
said difference. 

6. 10^Xl 2 =l(>2\cu. ft. 

7. .'. i^XlO^cu. ft.=5.277865652 cu. ft.=the volume. 

III. .-. The volume of the segment is 5.277875652 cu. ft. 

CASE II. 
When the base is perpendicular to the revolving axis. 

Formula. — V=±n/i 2 ( — J(od — 2/i), where d is the 
revolving axis, D the fixed axis, and h the height of the segment. 

Rule. — (1) Divide the fixed axis by the revolving axis, and 
multiply the quotient by the difference between three times the 
revolving axis and twice the height of the segment, 

(2). Multiply the product, thus found, by the square of the 
height of the segment, and this product again by ]r7t. 

I. Required the volume of the segment of a prolate spheroid, 
its height being 6 inches, and the axes 50 and 30 inches respect- 
ively. 

By formula, V=\nh*(J?) (Zd-2h)={7tx6 2 Q£) X 



264 



FINKEL'S SOLUTION BOOK. 




(3X30— 2x6)=2450.442267 cti. in. 

1. 50 in.=the transverse diameter, or A 
axis. 

2. 30 in.=the conjugate d'mmeter2AfO. 

3. |=50-r-30=the quotient of the trans 
verse diameter divided by the 
conjugate diameter. 

4. 78 in.=3x30 in. — 2x6in.=the difference between three 
1 times the conjugate or revolving axis, and twice the 

height of the segment. 

5. 130 in.=f X78 in.=the product of said quotient by said 
difference. 

6. 4680 cu. in=130x6 2 =the square of the height of the 
segment by said product. 

7. .'. i?rX4680 cu. in.=2450.442269 cu. in.=the volume of 
segment. 

III. .*. The volume of the segment is 2450.442269 cu. in. 



XIV. CONOIDS. 

1. A. Conoid is a solid formed by the revolution of a conic 
section about its axis. 

I. THE PARABOLIC CONOID. 

1. A JParabolic Conoid is the solid formed by revolving a 
parabola about its axis of abscissa. 

Prob. CXX. To find the surface of a parabolic conoid, or 
paraboloid. 

I dx 2 
Formulae. — (a) S= C27tyds== C27ty<s\\-{-— 2 dy= 

where 2ft is the latus rectum of the parabola and jy the radius of 
the base of the conoid, or the ordinate of the parabola. 

3 3 

(3) S—\n4ft\ (ft-\-x) — ft |, where 2ft is the same as above 

and x the altitude of the conoid, or the axis of abscissa of the 
parabola. 

Rule. — To the square of half the latus rectum, or ftrinciftal 
ftarameter , add the square of the radius of the base of the conoid 
and extract the square root of the cube of the sum; from this re- 
sult \ subtract the cube of half the latus rectum and multiftly the 



MENSURATION 



265 



difference oy2lt, and divide the product by one andone half times 
the latus rectum. 

I. Determine the convex surface of a paraboloid whose axis is 
20, and the diameter of whose base is 60. D 

From the equation of the parabola, y 2 =2px : 
we have 30 2 =2/X20 ; whence 2^=45. 

.-. By formula(^),^^|(/2_|_ r 2) f _ / 3| 

-& iKf )*+»•]<-(¥ ) 

^X25X(125— 27)=49X25X3.14159265= FIG. 52. 

3848.45118. 



1. 30= the radius A O of the base of the conoid. 

2. 20=the altitude OD. Then by a property of the para- 

bola, 

3. 30 2 =2/x20, whence 

4. /=22^, the principal parameter of the parabola. 




II. 



uare root of 



5. (y) 3 Xl25= > J[(22i) 2 +30 2 y=the sq 

the cube of the sum of the squares of half the latus 
rectum and the radius of the base. 

(45X 3 
•— 1 =the cube of half the latus rectum. 

7 CD 3 * 125 -Gf y^T^THthe differ- 

ence between said square root and the cube of half 
the latus rectum. 

8. 2*x(~y (125— 27)= 7 rX9Sx(^) 3 =2 7 r times 
said difference. 

9. .-. 2^98x('^y^-(|X45)=3S48.45118=thesurfaceof 
the conoid. 

III. .*. The surface of the conoid is 3848.45118. 

Prob. CXXI. To find the volume of a parabolic conoid. 

Formula. — V== C7ty 2 dx= f7r2pxdx=ftpx 2 = 
^7r(2px)x=^7ry 2 x, whereby is the radius of the base and x the 
altitude. 

Rule. — Multiply the area of the base by the altitude and take 
half the product. 



266 FINKEL'S SOLUTION BOOK. 

T. What is the volume of parabolic conoid, the radius of 
whose base is 10 feet and the altitude 14 feet? 

By formula, V=\7t y 2 x = \n\W X 14=700x^=2202.114855 
cu. ft. 

1, 10 ft.=the radius of the base. 

2. 14 ft.=the altitude. 

\\.\ 3. 7rl0 2 =314.159265 sq. ft. the area of the base. 

4. .-. ix 14X314.159265=2202.114855 cu. ft.=the volume 
^ of the conoid. 

III. .-. The volume of the conoid=2202. 114855 cu. ft. 
Note.— Since the volume of the conoid is ±TTy 2 x, it is half of its circum- 
scribed cylinder. 

Prob. CXXII. To find the convex surface of a frustum of 
a parabolic conoid of which the radius of the lower base is K 
and the upper base r. 

r R 2n c a | ) 

Formula.— S= I 27tyds=-^ (pz+Riy—^+r*) 2 £ . 

I. What is the volume of the frustum of a parabolic conoid of 
which the radius of the lower base is 12 feet, the radius of the 
upper base 8 feet, and the altitude of the frustum 5 feet? 

Since 12 2 =2^*' and 8 2 =2/*, 12 2 — 8 2 =2/(*'— x). Bnt x'— x 
=5 feet. .■'. 12 2 —8 2 =2/Xo, whence 2/=16, the latus rectum. 

... By formula, S= g[(^ + ^) f _(^ + r 2 ) ^"J_ 
^[(8 2 +12 2 ) l -(8 2 +8 2 ) l ]=^(832v'i3-1024V2)= 

1 6 
3 



413^13—16^2]. 



Prob. CXXIII. To find the volume of the frustum of a 
parabolic conoid, when the bases are perpendicular to the 
axis of abscissa. 

Formula. — V=^7tR 2 x / — ±7rr 2 x=±7r(x / — x)(R 2 -\-r 2 ) 
%7ta(R 2 +r 2 ). 

Rule. — Multiply the sum of the squares of the radii of the 
two bases by n and this product by half the altitude. 

I. What is the volume of the frustum of a parabolic conoid, 
the diameter of the greater end being 60 feet, and that of the 
lesser end 48 feet, and the distance of the ends 18 feet? 

By formula, \ r ^7ta(R 2 -\-r 2 )=|X18X^(30 2 +24 2 )=9tt(900 
_[-576)=9Xl476X7r==132847r=41732.9177626 cu. ft, 



MENSURATION. 



267 



"1. 30 ft.=the radius of the larger base. 

2. 24 ft.=the radius of the lesser base. 

3. 18 ft.=the altitude of the frustum. 

4. 900 sq. ft.=the square of the radius of the lower base. 

5. 576 sq. ft.=the square of the radius of the upper base. 

6. 1476 sq. ft =900 sq. ft.+576 sq. ft —their sum. 

7. .-. iX 18 XttX 1476=13284 x*=41732.9177626 cu. ft.= 
the volume ot the frustum of the conoid. 

.-. The volume of the frustum is 41732.9177626 cu. ft. 

II. THE HYPERBOLIC CONOID. 

1. An Hyperbolic Conoid is the solid formed by revolv- 
ing an hyperbola about its axis of abscissa. 

Prob. CXXIV. To find the surface of an hyperbolic con- 
oid, or hyperboloid. 



IlJ 



III. 



Formula.— S=J2nyds=2 n J>>] 1 -\-( l -j- J 



d.X: 



*/>j 



dx- 



dx- 



a t 



a- 



"In I -*Je 2 x 2 — a 
J a 

log[>-t-i\W 2 — a~ 2 ] I -f-C=7r- \ X sle e < 



x\c-x- — a 1 — 



a( 



V« 2 +^ : 



loof 



a-\ 



ab 



*/a*+6' 



: J r ^sJ e 2 x 2_ l 



Prob. CXXV. To find the volume of an hyperbolic conoid. 

Formula. — V=^7t(R 2 -\-d 2 )k i where R is the radius of 
the base, d the middle diameter, and h the altitude. 



ill 



Rule. — To the square of the radius of the base add the square 
of the middle diameter between the base and the vertex; and this 
sum multiplied by the altitude, and the product again by \tt, 
give the solidity- 

I. In the hyperboloid A CB, the altitude 
CO is 10, the radius A O of the base 12, 
and the middle diameter DE 15.8745 ; 
what is the volume ? 

fl. 10=the altitude CO. 

2. 12=the radius A O of the base. 

3. 15.8745=fhe middle diameter DE. 

4. 144=12 2 =the square of the radius 
of the base. ™' 53 ' 

the square of the middle diameter. 



of %£ 



268 FtNK£L>S SOLUTION BOOK. 

6. 395.99975=251.99975+144==the sum of the squares of 
the radius of the base and the middle diameter. 

7. .-. \?tx 10 X 395.99975=2073.45469 l=the volume. 

III. „•. The volume of the conoid is 2073.454691. 

Prob. CXXVI To find the volume of the frustum of an 
hyperbolic conoid. 

Formula.— V=\rta(R' 1 -\-d 2 -\-r 2 ), where R is the ra- 
dius of the larger base, and r the radius of the lesser base, and d 
the middle diameter of the frustum. 

Rule. — Add together the squares of the greater and lesser 
semi-diameters , and the square of the whole diameter in the mid- 
dle', then this sum being ?nultiplied by the altitude, and the prod- 
uct again by \7t, will give the solidity. 

XV. QUAD^ATUI^E AND CUBATUI^E OF 
SURFACES ANTE) SOLIDS OF REVOLU- 
TION- 

1. CYCLOID. 

Prob. CXXVII. To find the surface generated by the 
revolution of a cycloid about its base. 

Formula.— S=2J2 nyds=±n Jfldx*+dy*== 

Jo \2r—y 
Rule. — Multiply the area of the generating circle by ^. 



Prob. CXXVIII. To find the volume of the solid formed 
by revolving- the cycloid about its base. 

f2r y z dy 
Formula. — V=2 f7ry 2 dx=27r I , J ==57r 2 r 3 = 

J Jo V2ry—y~ 

iX7r(2r) 2 x27tr. 

Rule. — Multiply the cube of the radius of the generating cir- 
cle by 57T 2 . 

Prob. CXXIX. To find the surface generated by revolv- 
ing- the cycloid about its axis. 

Formula.— S==J27ryds=A7t)/2?Jy—=8ttr 2 (7r—i). 

Rule. — Multiply eight times the area of the generating circle 
by n minus \. 

Prob. CXXX. To find the volume of the solid formed by 
revolving the cycloid about its axis, 




MENSURATION. 269 

Formulu. — V= f7ry 2 dx=27t Cy%(2r — y)~~\dy^. 

Rule. — Multiply \ of the volume of a sphere whose radius is 
that of the generating circle by \rt 2 — f. 

Prob. CXXXI. To find the surface formed by revolving 
the cycloid about a tangent at the vertex. 

Let P be a point on the curve, A£=P£=y, J5P=A£=x, 
A C= CF=r,and the angle A CF 
=0. Then we shall have AB= 
y = AC — CE=r — r cos 6 ; and 
AB = x=FP+EF=AF+EF 
=r0-\-r sin 0. 

.-. Formula. — S= 

Art Cy\dx 2 -\-dy 2 =47tr I (r — fig. 54. 

rcos#vV 2 (l+cos#) 2 +r 2 sin 2 6 dd = 4 n r 2 f* (l_cos9)X 

Jo 

\ / 2+2cos^^=87rr 2 } (l—co*0)cos%fidd=lS7rr 2 I * (1— C os s £0) 

Jo Jo 

Xcos|ft#?=167rr 2 J (cosi6—cos*%6)d0=167rr* [~2sin^— 

tsin-Jflcos 2 -^— £sin|0~] *=Af nr*. 

— to 
Rule. — Multiply the area of the generating circle by "\ 2 . 

Prob. CXXXI1 To find the volume formed by revolving 
a cycloid about a tangent at the vertex. 

Formula.— V'^Ifny 2 dx=&rt j (r — rcos0) 2 r(l-}- 

cos0)dd=27rr* j (\—cos0) 2 (\-\-cosO)d0=2nr* /^(l—cosfl— 

Jo Jo 

cos 2 0-{-cos 3 0)d0= 7rV 3 =the volume generated between the 
curve and the tangent. 

.-. V=7zAD*xGH—V'=7r(2r) 2 x27zr—7i°rZ=77r 2 rZ. 

Rule. — Multiply the cube of the radius of the generating cir~ 
cleby7n 2 . 

2. CISSOID. 

Prob. CXXXI 1 1. To find the volume generated by revolv- 
ing the cissoid about the axis of abscissa. 

x 3 
Formula. — V= C ny^dx= Cn- dx=7t( — J* 3 — ax 2 

9/ 



_-4a** +8o «log(^). 



270 FINKEL'S SOLUTION BOOK. 

Prob. CXXXIV. To find the volume formed by revolving 
the cissoid about its asymptote. 

Formula.— V=2jn(AR) 2 dy(Fig. 22)=2n: (2a— x) x 

( 3 ^~*) **dx=27r[j(2ax— x 2 )%-\-2aJ(2ax— x 2 )dx\=2n 2 a*. 
y Let — xy% 

Prob. CXXXV. To find the volume formed by revolving* 
the Witch of Agnesi about its asymptote. 

Formula. — V— I 7ty 2 dx=\ ny^x—^na C(2ay — 

y2)\dy\ =±n*a* 

Prob. CXXXVI. To find the volume formed by revolving 
the Conchoid of Kicomedes about its asymptote, or axis of 
abscissa. 

Formula.— V=j7ty^dx=n /T— ^!_ a . +^(^ 8 — j/ 2 )* 

2. SPINDLES. 

A. Circular Spindle is the solid formed by revolving the 
segment of a circle about its chord. 

Prob. CXXXVII . To find the volume of a circular spindle. 

Let AEBD be the circular spindle formed 
by revolving the segment A CBE about the 
chord A CB. Let AB=2a, the length of 
the spindle, and ED==2d, the middle diame- 
ter of the spindle. Let CI=KL=x, the ra- 
dius of any right section of the spindle, and 
ICI=CB=y. Then the required volume of F/G. 5™ 

the spindle is V=2tt j x 2 dy. . .(1). Let R=( a 2 -\-b 2 )-±-2b . .(2), 
«/ o 

be the radius of the circle and 6 the angle A GE. Then by 
a property of the circle, KI 2 =(2R—EI)xEI, or y 2 =(2R— 
EI) x EI. But EI=E G—IG=R—(IC-\- CG)=R— 

(x+RcosO). ' .-. y*=\2R— [R— (x+Rcosd) 1}r—( X -\-Rcos0)1 

=[R+(Rco sV+x)]X [I?—(I?cosO+x)]=R2—(Rcos0+x) 2 ; 
whence x=*lR 2 — y 2 — Rcosd . . (3). Substituting this value of x 

in (1), we have V=2n \ (slR 2 —y 2 —Rzosd) 2 dy=2n\R 2 (\+ 
Jo 




COS 



MENSURATION. 271 

e )y— \y Z — 2^?cos6'Ai? 2 sin ^~ iys'X —y 2 \V=27T j %a* 

Rllle. — Multiply the area of the generating segment by the 
path of its center of gravity. — Guildiri's Rule. 

3. THE PARABOLIC SPINDLE. 

A Parabolic Spindle is a solid formed by revolving a 
parabola about a double ordinate perpendicular to the axis. 

Prob. CXXXVIII. To find the volume of a parabolic 
spindle. 

Formula.— V=2 f 7t(k—x) 2 dy=27t f (h 2 —2hx+ 

J o Jo 

x*)dy=2n£(Kdy-2^ 

y*l =27rh 2 y— %/ixy+lx 2 yl =27r]h n ~b—%hbx+lbx 2 ]. But x= 

h y whenjy=£. 

.-. V=2n[/iH— %hH+lh*b]=\<l 7 rh' 2 b== T Kx2bXKh~. 

Rule. — Multply the volume of its circumscribed cylinder by r 8 5 . 

I. What is the volume of a parabolic spindle whose length 
A C is 3 feet and height BD 1 foot? 

By formula, 1=\ in h-b= /Wx 1- X :i=4.9945484 cu. ft. 
1- 1 ft.=height BD of the 
spindle. 

2. 3 ft —length A C. 

3. 7T Xl 2 X 3=9.42477795 cu. ft. 
the volume of its circum- 
scribed cylinder. 

4. .-. &X9.42477795 cu. ft.= 
4.9945484 cu. ft., the vol- 
ume of the parabolic spin- 
dle. FIG. 56. 

III. .". The volume of the spindle is 4.9945484 cu. ft. 

Prob. CXXXIX. To find the volume generated by revolv- 
ing the arc of a parabola about the tangent at its vertex. 

Let APC be an arc of a parabola revolved about AB, and let 
P be any point of the curve. Let AE=PF=x, and AJF=PP 
==y. Then the area of the circle described bythe line PF is nx 2 . 

iji d y =2nx %tf2 x 



ii. 





FIG. 57. 



272 FINKEL'S SOLUTION BOOK. 

*y 5 —^irx 2 y=^Tfk 2 i>, where /?— the height, and b== CD, the or- 
dinate of the curve. 

Rule. — Multiply the volume of its circum- 
scribed cylinder by \. 

Prob. CXIi. To find the volume generated 
by revolving- the arc APC of the parabola 
about BC parallel to the axis AD. 

The area of the circle generated by the line 
GP is 7r(b—y)\ 

.-. Formula. — V=n f(&— y) 2 dx=\nb 2 h. 

Rule. — Multiply the volume of its circumscribed cylinder by \. 

Note. — In the last two problems, the volume considered, lies between 
the curve and the lines AB and B C respectively. The volume generated 
by the segment A CD is found by subtracting the volume found in the two 
problems from the volume of the circumscribed cylinders. 

Prob. CXLJ. To find the volume formed by revolving- a 
semi-circle about a tangent parallel to its diameter. 

Let the semi-circle be revolved about the tangent A C. Let 
AC=R,PP=AG=BC=y,AP=GP=x. Then 
the area of the circle generated by the line GP is 
nx 2 . But x 2 =2R 2 — 2R(R 2 — y 2 )%— y 2 \ for, 
FC 2 =P C 2 —PP 2 ,or (R—x) 2 =R 2 —y 2 ; whence 
x=R—\ / R 2 —y 2 , and x 2 =2R 2 —2R*jR 2 —y 2 —y 2 

.-. Formula.— V=2j7tx 2 dy=27t J(2R 2 — 
2R\ i R 2 ^ 2 —y 2 )dy=\7t R* ( 10— 3 tt), which is the 

entire volume external to the semi-circle- 

FIG. 59. 
Rule. — Multiply one-fourth of the volume of a sphere zvhose 
radius is that of the generating semi-circle by (10 — Sn). 

XVI. I^EGULAI^ SOLIDS. 

1. A Hegular Solid is a solid contained under a certain 
number of similar and equal plane figures. 

2. Hie Tetrahedron, or Megular Pyramid, is a 

regular solid bounded by four triangular faces. 

3. Tlie Hexahedron, or Cube, is a regular solid bounded 
by six square faces. 

4c. The Octahedron is a regular solid bounded by eight 
triangular faces. 

5. The Dodecahedron is a regular solid bounded by 
twelve pentagonal faces. 




MENSURATION. 



273 



6. The Icosahedron is a regular solid bounded by twenty 
equilateral triangular faces. 

These are the only regular solids that can possibly be formed. 

If the following figures are made of pasteboard, and the dotted 
lines cut half through, so that the parts may be turned up and 
glued, together, they will represent the five regular solids. 




FIG. 59. 

1. TETRAHEDRON. 

Prob. CXLII. To find the surface of a tetrahedron. 

Formula. — S—l 2 VS, where / is the length of a linear 
side. 

Rltle. — Multiply the square of a linear side by *J]}=1.7820 
508. 

I. What is the surface of a tetrahedron whose linear edge is 
2 inches. 

By formula, S=/ 2 \ / 3=2 2 \ / 3=4\ / 3=6.9282 sq. in. 

{1. 2 in.=the length of a linear side. 
2. 4 sq. in.=2 2 =the square of a linear side. [surface. 

3. .-. V^X 4 sq. m.=1.73205x4 sq. in.=6.9282 sq. in., the 
III. .*. The surface of the tetrahedron is 6.9282 sq. in. 

Prob. CXLIII. To find the volume of a tetrahedron. 

Formula.— V=^l% / 3 , where /is the length of a linear 
side. 



274 FINKEL'S SOLUTION BOOK. 

Rule. — Multiply the cube of a linear side by T ^V2, or .11785. 

I. Required the solidity of a tetrahedron whose linear side 
is 6 feet? 

By formula, V*=frl2 / 3 = T V\ / 2X6 3 =18V2=25.455843 cu. ft. 

( 1. 6 ft.=the length of a linear side. 

II. < 2. 216 cu. ft.=the cube of the linear side. 

( 3, .-. T 1 2V / 2 _ X216 cu. ft.=V2xl8 cu. ft.=25.45843 cu. ft. 

III. .-. The volume of the tetrahedron is 25.45843 cu. ft. 



2. OCTAHEDRON. 
ProT). CXLJV. To find the surface of an octahedron. 
Formula.— 5=2^3 I 2 . 



Rule. — Multiply the sqttare of a linear side by2v3, i. <?., by 
two times the square root of three. 

I. What is the surface of an octahedron whose linear side is 
4 feet? 

By forumla, 5=2^ / 2 =2V / 3X^ 2 =32\ / 3=55.4256 cu. ft. 

1. 4 ft.=the length of a linear side. 

2. 16 sq. ft.=4 2 =the square of the linear side. 

XI n3. .-. 2V3X16 sq. ft.=V3x32 sq. ft.=l. 73205x32 sq. ft — 

55.4256 sq. ft. 
III. .-. The surface of the octahedron is 554256 sq. ft. 

Proh. CLXV. To find the volume of an octahedron. 
Formula.— V=±\/2P 

Rule. — Multiply the cube of a linear side by ^V#, I. e., by one- 
third 'of the square root of two. 

I. What is the volume of an octahedron whose linear side is 
8 inches? 

By formula, V=tffo /3 == xv / 2x8 3 =.47 140 15x512=241.359104 
cu. iii. 

1. 8 in.=the length, of a linear side. 

2. 512 cu. in.=8 3 =the cube of a linear side. 

3. .-.^2X512 cu. in =4X1-4142135X512 cu. in,= 
241.359104 cu, in. 



II. 



III. .*. The volume of the octahedron is 241.359104 cu. iu. 

3. DODECAHEDRON. 
Profo. CXLVI. To find the surface of a dodecahedron. 



MENSURATION. 



Formula.— 6=15^(^±^)/ 2 =20.6457 



285X/ 2 - 



Rule. — Multiply the square of a linear side by loV [i(£+ 
2V5)], or 20.64.57285. 

[. What is the surface of a dodecahedron whose linear side is 
3 feet? 



X9 



II 



By formula, 5=15'\I (^5i^-^.^/2=20.6457285 

185.8115565 sq. ft. 

r l. 3 ft.=the length of a linear side. 

2. 9 sq. ft.=3 2 =square of a linear side. 

3. .-. 15^( 5 + 2Vo )x9 sq. ft=20.645728oX9 sq. ft. 
=185.8115565 sq. ft. 

III. The surface of the dodecahedron is 185.8115565 sq. ft. 

Prob. CXLVII. To find the volume of a dodecahedron. 
Formula.-^ F=5 > ](^^-^)/ 8 ==7.663115x/ 8 . 

Rule. — Multiply the cube of a linear side by 5 <>\i — — — 1, 

or 7.663115. 

I. The linear side of a dodecahedron is 2 feet ; what is its 
volume? 

By formula, F=5^(^|^)/ 3 =7.663115x8 
=61.20492 cu. ft. 

rl. 2 ft.=the length of a linear side. 
)2. 8 cu. ft.=2 2 =cube of a linear side. 
IL i3. .-. 5^(47+21^5)] X8cu. ft.=7.663115x8 cu. ft. 
I =61.20492 cu. ft., the volume. 

III. .'. The volume of the dodecahedron is 61.20492 cu. ft. 



4. ICOSAHEDRON. 

Prob. CXLVII I. To find the surface of an icosahedron. 

Formu la.— S=5V3/ 2 =8.66025 X l 2 . 

Rule. — Multiply the square of a linear side by 5\8, or 
8.66025. 



27G 



FINKEL'S SOLUTION BOOK. 



I. What is the surface of an icosahedron whose linear side 
5 feet. 

By formula, £=5VoY 2 :=5V3x5 2 :=125v1*==216.50625 sq. ft. 

rl. 5 ft.=length of a linear side, 
-ry J 2. 25 sq. ft.=5 2 =the square of a linear side. 



,'o 



] 3. ••• 5V3X25 sq. ft.=S.66025x25 sq. ft.=216.50625 sq. ft. 
j, =the surface. 

III. .*. The surface of the icosahedron is 216.5062o sq. ft. 

Prob. CXLIX. To find the solidity of an icosahedron. 

formula.— K=f^[|(7+3V5)]/ 3 =2.18160x/ 3 ^ 

Rule. — Multiply the cube of a linear side ly fV[£(7-|-#V5)], 
or 2.18169 

I. What is the volume of an icosahedron whose linear side is 
3 feet? 

By formula, T=-gV[|(7+3V5)]/ 3 =2.181G9x3 3 ==58.90563cu. ft. 

rl. 3 ft.=the length of a linear side. 



II.; 



\2. 21 



cu. ft.=3 3 =the cube of a linear sid< 



3 - ■•• fv[i(7+3V5)]X27cu. ft.=21.8169x27cu. ft. 
I =58 90563 cu. ft.=the volume. 
III. .*. The volume of the icosahedron is 58.905G3 cu. ft. 
Note. — The surface and volume of any of the five regular sol- 
ids may be found as follows : 

Utile ( 1 ). — Multiply the tabular area by the square of a linear 
side, and the product will be the surface 

Rule (2). — Multiply the tabular volume by the cube of a 
linear side, and the prodztct will be the volume. 

Surfaces and volumes of the regular solids, the edge being 1. 



NO. OF 
SIDES. 


NAMES. 


SURFACES. 


VOLUMES. 


4 


Tetrahedron 


1.73205 


0.11785 


6 


Hexahedron 


6.00000 


100000 


8 


Octahedron 


3.46410 


0.47140 


12 


Dodecahedron 


20.64578 


7.66312 


20 


Icosahedron 


8.66025 


2.18169 



XVII. PRISMATOID. 

1. A JPvismatoid is a polyhedron whose bases are any two 
polygons in parallel planes, and whose lateral faces are triangles 
determined by so joining the vertices of these bases, that each 
lateral edge, with the preceding, forms a triangle with one side 
of either base. 



MENSURATION. . 277 

£. A Frlsmoid is a prisrnatoid whose bases have the same 
number of sides, and every corresponding pair parallel. 

Prob. CL. To find the volume of any prisrnatoid. 

Formula («).— V=±a(B 1 ^-3Az a )=±a(B 2 -\-3A / Za ), where 
a is the altitude, B. the area of the lower base, A Zn the area of a 

1 3 « 

section distant from the lower base two-thirds the altitude, B 2 
the area of the upper base, and A\ a the area of a section distant 
two-thirds the altitude from the upper base. 

Remark. — This simplest Prismoidal Formula is due to Prof. 
George 13. Halstead, A.M., Ph. D., Professor of Mathematics in 
the University of Texas, Austin, Texas, who was the first to 
demonstrate this important truth. The formula universally ap- 
plies to all prisms and cylinders; also to all solids uniformly 
twisted, e. g. the square screw; also to the paraboloid, the right 
circular cone, the frustum of a paraboloid, the hyperboloid of one 
nappe, the sphere,* prolate spheroid, oblate spheroid, frustum of 
a right cone, or ofa sphere, spheroid, or the elliptic paraboloid, 
the groin, hyperboloid, or their frustums. For a complete 
demonstration of the Prismoidal Formula, see Halstead" 1 s Elements 
of Geometry or Halstead 's Mensuration. 

Rule. — (a) Multiply one-fourth its altitude by the sum of one 
base and three times a. section distant from that base two-thirds 
the altittcde. 

Formula. — V=^a(B t -\-4:M-\-B 2 ) , where a is the alti- 
tude, B ' and B 2 the areas of the lower and upper bases respect- 
ively, and A/ the area of a section midway between the two 
bases. 

Rule. — (&) Add the area of the two bases and four times the 
mid cross-section; multiply this sum by one-sixth the altitude. 

XVII I. CYLINDRIC ICINGS. 

1. A Cylhldric Ring is a solid generated by a circle 
lying wholly on the same side of a line in 
its own plane and revolving about that line. 
Thus, if a circle whose center is O be re- 
volved about DC as an axis, it will gener- 
ate a cylindric ring whose diameter is AB 
and inner diameter 2 BC. OC will be the 
radius of the path of the center O. 

FIG. 60. 
Prob. CLI. To find the area of the surface ofa solid ring-. 

Formula. — S=&nrYjlnR=±7Z 2 rR, whare r is the ra- 
dius of the ring, and R is the distance from the center of the ring 
to the center of the inclosed space. 




278 



FINKEL'S SOLUTION BOOK. 



Rule. — Multiply the generating circumference by the path of 
its center. Or, to the thickness of the ring add the inner diame- 
ter and this sum being 7?iultiplied by the thickness, and the pro- 
duct again by 9.869704.4. will give the area of the surface. 

I. What is the area of the surface of a ring whose diameter 
is 3 inches and the inner diameter 12 
inches. 

By formula, S=\n 2 rR=\ 7 t l XHX 
(1-H-6)=tt 2 x45=9.8696044x45 
=444.132198 sq. in. 

'1. 1-J- in.=-J of 3 in.=the radius r 
of the ring. 

2. 6 in.=-J of 12 in.=the radius of 
the inclosed space. 

3. 6 in.-f 1-J- in.==7i in. = the ra- 
dius R of the center of the 



II. 




FIG. 61. 



ring. 

4. 7t A C=7t 3=the circumference of a section. 

5. 7t7K=27tIO=27r7^=7r 15=the path of the center. 

6. .-. 7r3X7rl5=7r 2 45=444.132198 sq. in.=tne area of the 
surface of the ring. 

III. .*. The area of the surface of the ring is 444.132198 sq. in. 

Prolb. CL/II. To find the volume of a cylindric ring. 

Formula. — V=7t 2 r 2 R=n r 2 X nR, where r is the ra- 
dius AI of the ring, and R the distance from the center of the 
ring to the center of the inclosed space. 

Rule. — Multiply the area of the generating circle by the path 
of its center. Or, to the thickness of the riitg add the inner di- 
ameter, and this sum being multiplied' by the square of half the 
thickness, and the product again by 9.8696044, 'will give the 
volume. 

I. What is the volume of an anchorring whose inner diame- 
ter is 8 inches, and thickness in metal 3 inches? 

By formula, V=7t 2 r 2 R=7r 2 X(H) 2 X(3+8)=24.75X 
9.8696044=244.2727089 cu. in. 

1. 1-J in.=-| of 3 in.=the radius of the ring. 

2. 8 in.=the inner diameter. 

3. 4 in.-j-1-J- in.=5^ in.=the radius R of the path of its 
center. 

* 4. 7r(l-J) 2 =the area of the generating circle. 

5. 27t(b±)=7t Xll=the path of its center. 

6. .-. 7rllx<ll) 2 =^ 2 X24.75=9.86044x24.75 
=244.2727089 cu. in., the volume of the ring. 

11L .'. The volume of the ring is 244.2727089 cu. in. 



II. 



MENSURATION. 279 

THEOREM OF PAPPUS. 

If a plane curve lies wholly on one side of a line in its own 
plane, and revolving about that line as an axis, it generates 
thereby a surface of revolution, the area of which is equal to the 
product of the length of the revolving line into the path of its 
center of mass ; and a solid the volume of which is equal to the 
revolving area into the length of the path described by its center 
of mass. 

XIX. MISCELLANEOUS MEASURE- 
MENTS. 

1. MASONS' AND BRICKLAYERS' WORK. 

JMciSOflS 9 work is sometimes measured by the cubic foot, 
and sometimes by the perch. A perch is 16^ ft. long, 1-^ ft. wide, 
1 ft. deep, and contains 16|XHXl=24| cu. ft. 

Prob. CLIII. To find the number of perch in a piece of 
masonry. 

Rule. — Find the solidity of the wall in cubic feet by the rules 
given for the mensuration of solids, and divide the product by 2^. 

1. What is the cost of laying a wall 20 feet long, 7 ft. 9 in. 
high, and 2 feet thick, at 75 cts. a perch. 
(1. 20 ft —the length of the wall, 

2. 7 ft. 9 in.=7| ft.=the height of the wall, and 

3. 2 ft.=the thickness. 

4. .-. 20X74X2=310 cu. ft.=the solidity of the wall. 

5. 24f cu. ft.= l perch. 

6. 310 cu. ft.=310-r-24|=124t perches. 

7. 75 cts. =the cost of laying 1 perch. 

8. .-. 12ffX75cts.=$9.39if=thecostof laying 124f perches. 
III. .-. It will cost $9.39if to layl2|| perches at 75 cts. a 

perch. 

2. GUAGING. 

€rdtl(/iflf/ is finding the contents of a vessel, in bushels, 
gallons, or barrels. 

' Prob. CLIV. To g"aug:e any vessel. 

Rule. — Find its solidity in cubic feet by rules already given ; 
this multiplied by 1728-^-2150.42 or .83, will give the contents in 
bushels; by 1728-^-231. will give it in wine gallons, which divided 
by 31\ will give the contents in barrels. 

Prob. CLV. To find the contents in gallons of a cask or 
barrel. 

Rule. — (1) When the staves are straight from the bung to 
each end; consider the cask two equal frustums of equal cones, 
and find its contc?its by the rule of Prob. XCIII, 



II. 



280 



FINKEL'S SOLUTION BOOK 



(2). When the staves are curved; Add to the head diameter 
{inside) two-tenths of the difference between the head and bung 
diameter; but if the staves are only slightly curved, add six- 
tenths of this difference; this gives trie mean diameter; express 
it in inches, square it, multiply it by the length i?i inches, and this 
product by .003 J + ; the product will be the contents in wine gallons. 

3. LUMBER MEASURE. 

Prob. CLVI. To find the amount of square-edged incli 
boards that can be sawed from a round log. 

Doyle's Utile. — Fro??z the diameter in inches subtract 
four; the square of the remainder will be the 7iumber of square 
feet of inch boards yielded by a log 16 feet long. 

I. How much square-edged inch lumber can be cut from; a 
log 32 in. in diameter, and 12 feet long? 

1. 32 in.=the diameter of the log. 

2. 12 ft.=the length, 

3. 32 in. — 4 in.=28 in.=the diameter less 4. 

4. 844 ft.=28 2 =the square of the diameter less 4, which 
by the rule, is the number of feet in a log 16 ft. long. 

5. 12 ft.=f of 16 ft. 

6. .-. | of 844 ft.=633 ft.=the number of feet of square- 
edged inch lumber that can be cut from the log. 

.*. The number of square-edged inch lumber that can be 
cut from a round log 32 inches in diameter and 12 ft. long is 
633 ft. 



II. 



III. 



4. GRAIN AND HAY. 

Prob. CLiVII. To find tlie quantity of grain in a wagon 
bed or in a bin. 

Rule. — Multiply the contents in cubic feet by 1728-^-2150.42, 
or .83. 

I. How many bushels of shelled corn in a bin 40 feet long, 
16 feet wide and 10 feet high ? 

1. 40 ft.=the length of the bin. 

2. 16 ft.=the width of the bin, and 

3. 10 ft.=the height of the bin. 

4. .:. 40X16X10=6400 cu. ft.=the contents of the bin in 
cu. ft. 

.5. .'. 6400 X -83 bu.=5312 bu.=the contents of the bin in bu. 
III. .-. The bin will hold 5312 bu. of shelled corn. 



IIA 



Rule. — (i) For corn on the cob, deduct one-half for cob. 
(2) For corn not "shucked" deduct two-thirds for cob and 
shuck. 






UA 



MENSURATION. 28i 

I. How many bushels of corn on the cob will a wagon bed 
hold that is 10| feet long, 3-^- feet wide, and 2 feet deep? 

1. 10^ ft.=the length of the wagon bed, 

2. 3| ft=its width, and 

3. 2 ft.=its depth. [in cu. ft 

4. .-. 10JX3|X2=73| cu. ft.=contents of the wagon bed 

5. .'. 73| X-8 bu=58.8 bu.=n umber of bushels of shelled 
corn the bed will hold. 

6. .'. \ of' 58.8 bu=29.4 bu.=the number of bushels of 
corn on the cob that it will hold. 

III. .*. The wagon bed will hold 29.4 bu. of corn on the cob. 

Prob. CliVIII. To find the quantity of hay in a stack,rick, 
or mow, 

Rule. — Divide the cubical contents in feet by 550 for clover or 
by 450 for timothy; the quotie?it will be the number of tons. 

Prob. CLXIX. To find the volume of any irregular solid. 

Rule. — Immerse the solid in a vessel of water and determine 
the quantity of water displaced. 

I A being curious to know the solid contents of a brush 
pile, put the brush into a vat 16 feet long, 10 feet wide, and 
8 feet deep and containing 5 feet of water. He found, after 
putting in the brush, that the water rose 1^ feet ; what was the 
contents of the brush pile? 

'1. 1G ft —the length of the vat, 

2. 10 ft.=the width, and 

3. 1^ ft.=the depth to which the water rose. 

4. .-. 1()X10XH=240 cu. ft.=the volume of the brush pile. 
III. .*. 240 cu. ft =the volume of the brush pile. 

XX. SOLUTIONS TO MISCELLANEOUS 

PROBLEMS. 

Prob. CLX. To find at what distance from either end, a 
trapezoid must be cut in two to have equal areas, the divid- 
ing line being parallel to the parallel sides. 

Fo rmula.— d=A-~[^(b 2 +bl)-{-b]^(b+b 1 )a 
~^~[^i(^ 2 +^i)+^]' wnere ^ is the area of the trapezoid, b the 
lower base, and b x , the upper base. ^(b~-\-b'l ) is the length of 
the dividing line. 

Rule. — 1. Extract the square root of half the sum of the 
squares of the parallel sides and the result will be the length of 
the dividing line. 



II. 



282 



FINKEL'S SOLUTION BOOK. 



2. Divide half the area of the whole trapezoid By half the sum 
of the dividing li?te and either c?zd, and the qitotient will be the 
distance of the dividing line from that e?id. 

I, I have an inch board 5 feet long, 17 inches wide at one 
end and 7 inches at the other; how tar from the 
large end must it be cut straight across so that 
the two parts shall be equal? 

By formula, d =\(b-\-b 1 )g- ^[V|(£ 2 -|-£;-)+4] 
= |(17+7)60-^[V|(17 2 +7 ¥ )+17]=720-:-30 
=24 in. =2 ft. 

r 1. 
2. 

3. 



IM 



10. 

11. 

12. 
13. 
14. 
15. 
16. 

17. 

IS. 




But 



Let ABCD be the board, [end, 

AB=17 in.=#, the width of the large 
DC=7 in.=y, the width of the small 

end, and [board. 

JfK=5 ft.=60in.=a, the length of ihe 
Produce HK, AD, and EC till they 

meet in E. Then by similar triangles, 
ABE\EGL\EDC\\AB^\LG"\DC <1 . 
BGL=E£>C+i(Al?C£>), or 
2E GZ=2ED C+AB CD=ED C+BZ> C-\-AB CD 

=EDC+EAB. 
.-. EGL=i(EDC+BAB), i. c.,EGL is 

tic mean between EAB and EDC. 
.-. GE 2 =i(AB^DC <2 )=i(d^y 2 )=an 

mean between EAB and ED C, 

GL=Ji(6*+l>' 2 )=iV2(£-+£ /2 ). 

Draw CM perpendicular to AB. 

FL=\ GL=^2{b' 1 ^b / ^) . 

IL=EZ—EI(=I£C=±D C=±b)-- 

CM=HK=a. 

MB=\{b—b'). Then in the similar triangles CMB 

and CVZ, 
MB\IL\\CM\CI, or \(b—V)\(^2(y-\-y~ )—, ]b)::a: 
CI. Whence 
CI=a(i\/2(b*+b'>)— |J)-H(J— *')=■ 

a(^2(b 2 +b^)- 



arithme- 



rithmetic 



xV2(^.+3' 2 )— \b. 



-3)_ e o (jV2(17^+7 2 )-7) 

17 — 7 



=3G in. 



b—y 

=3 ft. 
19. .-. IM=CM—CI=5 ft— 3 ft.=2 ft., the distance from 
the large end at which the board must be cut in two 
to have equal areas. 

III. .*. The board must be cut in two, at a distance of 2 feet 
from the large end, to have equal areas in both parts. 

(i?. H. A., p. 407,prob. 101.) 



MENSURATION. 



283 



Prob. CLXI. To divide a trapezoid into n equal parts and 
find the length, of each part. 

Formula— ^h=rzp | %]- *~ — £~| • 

*»^p->]( [ *~ ( *~ i)] r +( *~ i) 0] ■ where * is 

the width of the small end, b the width of the large end, and a 
the length of the trapezoid. h l is the length of the first part at 
the small end, k 2 the length of the second part, and so on. 

I. A board ABCD whose length BC is 36 inches, width 
AB 8 inches and DC 4: inches, is divided into three equal pieces. 
Find the length of each piece. 

By formula, A 1= =_ [^ Lj± 6l ]= 

¥ 3 _ 6 ¥ [Vi(3— l)4 2 -h8 2 — 4]=9[V32— 4]=36(V2— 1) =14.911686 in. 

=11.442114 in. A,^[j(^ 3 ^+ 3ft LJBSgP] 

=36[2— V3]=9.6462 in. 

1. 4in.=the width DC oi the small 
end, 

2. 8 in.=the width A B of the large end, 
and 

3. 36 in =the length BC of the board. 

4. .-. 216 sq. \n—\(AB+DC)xBC 
=£(8+4) XM=the area of the 
board. 

5. £ of 216 sq. in.=72 sq. in.=the area of 
each piece. 

6. AK=AB— KB(=DC)=Sin.— 4 in. 
=4 in. In the similar triangles 
AKD and DCB, 

7. AK\DK\\AB\BE, or 4 in.:36 in.::8 in. \BE. Whence, 

8. ^^=(36x8)-^-4=72 in. [triangle ABE. 
} 9. .-. |(^^X^^)=i(8X72)=288sq.in.=theareaofthe 
<10. ABE—ABC£>=288 sq. in.— 216 sq- in.=72 sq. in. 

=area of the triangle DCE. 




FIG. 63. 



284 



FINKEL'S SOLUTION BOOK. 



11. 



12. 



14. 
15. 



DCE-\-DCGF=72 sq. in.+72 sq. in. = 144 sq. in. 

=the area of the mangle EGE. 
DEC+DCGF+EGIH=12 sq. in.+72 sq. in.+72 

sq. in. =216 sq. in.==th'e area of the triangle HIE 
EEG:DEC\\EG*\EC*, or 

144 s q. in.:72sq. in.:: GE 2 'M 2 . Whence, 

G.£=V / (144x36 2 )-^72=36V'2=50.911686 inches. 
.-. GC=GE—CE=5Q$ 11686 in.— 36 in.=14.911686 
in., the length of EG CD. Again, 

16. DEC\HIE\\EC 2 \EI 2 , or 

72 sq. in.:216sq.in.::3 6 2 :^/ 2 . Whence, 

17. Ef=J (216 x36 2 )-t-72=36xV3=62.3538 in._ 

18. .-. GI=EI—EG=SQs/l—SQ^2=S6(V i 6—V2) 
=11.442114 in., the length oiHIGF, and 

19. BI=EB=EI=n— 36V§=36(2— V3)=9.6462 in., the 
length oZABIH. 

r ^/=9.6462 in., 
III. .-.{ £7=11.442114 in., and 
lGC=14.911686in. 

Proh. CL.XII. To find the edge of the largest cuhe that 
c an he cut from a sphere. 



Formula 



— ^=<\] 



Z> 2 ,- 

— =iV37>=. 57735 X^ 



D 



is the diameter of the sphere. 

Rule. — Divide the square of the diameter of the sphere by 
three and extract the square root of the quotient; or, ?nultiply the 
diameter by .57785. 

I. What is the edge of the largest cube that can be cut from a 
sphere 6 inches in diameter? 



By formula, e- 



\D 2 /— 

. v 36 



:6xVWV3x6=.57735x6 



=4.4641 in. 

A. 6 in.=the diameter of the sphere. 
II.<J2. .'. .57735X6 in.=3.4641 in.=the edge of the largest cube 
I that can be cut from the sphere. 

III. .*. The edge of the largest cube that can be cut from a 
sphere whose diameter is 6 inches, is 3.4641 in. 

Proh. CLXIII. To find the edge of the largest cuhe that 
can he cut from a hemisphere. 



Formula, 



- e=^=iVe>xD=. 



408248 XD. 



Rule. — Divide the square of the diameter by 6, a?id extract the 
square root of the quotient ; or, multiply the diameter by .1^082^8. 



MENSURATION. 285 

I. What is the edge of the largest cube that can be cut from a 
hemisphere, the diameter of whose base is 12 inches? 

By formula, e=V Z> 2 ~6=V / ^=12V'I=iV'6Xl2=.408248 
•X 12=4.899176 in. 

T y ( 1. 12 in.=the diameter of the base of the hemisphere. 

(2. .-. .408248X12 in.=4.899176 in. 
III. .*. The edge of the largest cube that can be cut from a 
hemisphere, the diameter of whose base is 12 feet, is 4.899176 in. 

Prob. CLXIV. To find the diameter or radius of tlie three 
largest equal circles that can he inscribed in a circle of a 
given diameter or radius. 

Formula.— d=D-±-( 1+f V / 3)=Z)~2.1557= .4641 X D 
or r =i?-^(l+tV3)=.4(541X^. 

K-ule. — Divide the diameter or radius of the given circle by 
2.1557 and the quotient will he the diameter or radius of the three 
largest equal circles inscribed in it; or, multiply the diameter or 
radius by .^6^1, a?id the result will be the diameter or radius re- 
spectively of the required circles. 

I. A circular lot 15 rods in diameter is to have three circular 
grass beds just touching each other and the larger boundary ; 
what must be the distance between their centers, and how much 
ground is left in the triangular space about the center? 

By formula, 2r=2/?^(l+fV3)=2/^2.1557=s. T V.rr 
=6.9615242 rd.=the distance between their centers. 

Construction. — Let AHE be the circular lot, C the center, and 
ACE any diameter. With E as a center and radius equal to 
CE describe an arc intersecting the circumference of the lot in H. 
Draw a tangent to the lot at E and produce the radius CH to 
intersect the tangent at B. Bisect the angle CBE and draw 
the bisector GB. It will meet the radius CE in G, the center 
of one of the grass beds. Draw GF perpendicular to CB. Then 
GF=GE, the radius of one of the grass beds. Draw EH. 
Then EH=CH=EC, and CH=HB, because the triangle 
EHB is .isosceles. 

1. CE=7i rd.=7?, the radius of the lot. 

2. CB=2 CH=2R. 

3. BB=V CB' 2 — CE 2 =V (*2R)-—R 2 =RVz. In the 
similar triangles CFG and CBE, 

4. CF:FG::CE::FB,ov CF:GF;:R:RVW. But 

5. CF=CB— FB(=FB)=2R— RVs=R(2— VI). 

6. .. R(2—VS):GF::R:RVS. Whence, 

7 GF=- 7 ^%=R(2V%—n)=7H2VS-S)=nx 

i—Vs 

.4641=3.48075 rd.=the radius. 



286 



FINKEL'S SOLUTION BOOK, 



1U 



GK=2r=2R(2VZ— 3)- 

tween their centers. 



:6.9615 rd., the distance be- 



3. 



GD=V GK*—DK*-- 
i(7KX Gp)=%(2r 

area of the triangle 

IGK. 

Area DKF=\ of the 

small circle, because 

the angle l)KF is 



60°, or 1 of 360°. 



^4. 

5. 



.;. Area DKF=\nr*. 
^7tr 2 =S times-i-7rr 2 




:the area of the FIG. 64 

three parts of the small circles within the triangle 
IGK. 

rWE— |7rr 2 =r 2 (V3— £?r )=.16125368r 2 
=.16125368 X[/?(2V3— 3)] 2 =.16125368X(21 
— 12V / 3)i? 2 =.16125368*X-2153904x/^ 
=.03473265 x^ 2 =03473265x(7i) 2 =1.953712 
sq. rd.=the area of the space inclosed. 
f6.9615 rd.=the distance between their centers, and 
III. .'.-J 1.953712 sq. rd.=the area inclosed about the center of 



I the given lot. 



(R. J7.A..p. 407, prod. 100.) 



Prob. CLXV. Having- given the area inclosed by three 
equal circles to find the radius of a circle that will just in- 
close the three equal circles- 



NV.03473265y 



■3) 2 (V'3-i^) 
where A is the area inclosed. 



473265. 

Rule. — Divide the area inclosed by .034.73265 and extract the 
square root of the quotient, and the result will be the radius of the 
required circle. 

Prob. CLXVI. Having given the radius «,, &, c, of the 
three circles tangent to each other, t<» find the radius of a 
circle tangent to the three circles. 

Formula. — r or r / =— — — — — — , y , ,-, — j—r-. rr-\' 

2\/[abc(a-\-b+c)]^(ab+ac-\-bc) 

the minus sign giving the radius of a tangent circle circumscrib- 
ing the three given circles and the plus sign giving the radius 
of a tangent circle inclosed by the three given circles. 

Note.— This formula is due to Prof. E. B. Seitz, Late Professor of Mathe- 
matics in the North Missouri State Normal School, Kirksville , Mo., of 
whom we give a biographical sketch accompanied by his photograph. 

This fo»mula is taken from the School Visitor, Vol. II. p. 117, with the 



MENSURATION. 287 

slight change that the plus sign is introduced for the case in which the 
tangent circle is inclosed by the three given circles. The problem of finding 
two circles tangent to three mutually tangent circles, is one supposed 
to have been proposed by Archimedes more than 2000 years ago, though 
the problem he proposed was not so general — the diameter of one of the 
given circles being equal to the sum of the diameters of the other two 

The problem of finding all circles that can be drawn within three mu- 
tually tangent circles and tangent to each of them, has been simply and 
elegantly solved by D. H. Davison, Minonk, 111. The above formula led 
him to his wonderful solution. For a complete and elegant solution, 
where he has actually computed and constructed 81 circles tangent to three 
given circles, see School Visitor, Vol. VI , p. 80. 

Prob. CLXVII. To find the surface common to two equal 
circular cylinders whose axes intersect at right angles. 

Formula. — S=16R 2 , where R is the radius of the cylinders. 

Rule. — Multiply the square of the radius of the intersecting 
cylinders by J6. 

I. If the radius of two equal circular cylinders, intersecting 
at right angles is 4 feet, what is the surface common to both? 
By formula, S= 1QR 2 = 16 X4 2 = 256sq. ft. 

4 ft. = the radius of the cylinders. 
11.^ 2. 16 sq. ft. = 4 2 = the square of the radius of the cylinders 
.*. 16 X 16 sq. ft. = 256 sq. ft. = the surface common to the 
. two cylinders. 

III. .*. 256 sq.ft. = the surface common to the two cylinders. 

Prob. CLXVII I. To find the volume common to two equal 
circular cylinders whose axes intersect at right angles. 

Formula. — V== ^Z? 3 , where R is the radius of the 
cylinder. 

Rule. — Multiply the cube of the radius of the cylinders 
by tf. 

1. A man digging a well 3 feet in diameter, came to a log 3 
feet in diameter lying directly across the entire well; what was 
the volume of the part of the log removed ? 

By formula, V= ^i? 3 = ^(f) 8 =18 cu. ft 

1. 3 ft. = the diameter of the log and the well. 



II. 



2. 1 ^ ft. = the radius. 



& 



3. 3| cu. ft. = ( 1 1) 3 = the cube of the radius. 
I 4. .'. *f X 3f cu. ft.= IScu.ft. , the volume of the part of 
I the log removed. 

III. .'. The volume of the part of the log removed is 18 cu.ft. 

Prob. CLXIX. To find the height of an object on the 
earth's surface by knowing" its distance, the top of the ob- 
ject being 1 visible above the horizon. 



288 FINKEL'S SOLUTION BOOK. 

L,etBF=ahe any object, AB—t a tangent to the earth's sur- 
face from the top of the object, and FE=D the diameter of the 
earth. Then by Geometry, AB 2 =BF( 7? F-\-FB), or t 2 =a(a 

t 2 
-\-D). .'. a= ■ _ ■■ But a is very small as compared with the 
a-\-D 

diameter of the earth and AB—AF without appreciable error. 

AF 2 c 2 
.'. Formula. — a— — =r — =— , where c is the distance to 

the object from the point of observation. 

When c=l mile, a= — — — =-| ft., nearly. 
7912 3 J 



\/e 



: /g 



By formula, «=_=__ = __ X 5280=|X 10 2 ==66f ft. 



Rule. — Multiply the square of the distance in 
miles by%, and the result will be the height of the 
object in feet* 

I. What is the height of a steeple whose top can 

be seen at a distance of 10 miles? FIG. 65. 

c 2 _ 1Q 2 _ 1Q2 

-Z> — 7912~7912 

rl. 10 miles=the distance to the steeple. 

II.<2. 100=10 2 =the square of the distance. 

13. .-. f of 100--= 66f ft.=the height of the steeple. 
.III. .-. The height of the steeple is 66f ft. 

Prob. CLXX. To find the distance to an object by know- 
ing its height, the top only of the object being- visible above 
the horizon. 

Rule. — Multiply the height of the object infect by § and ex- 
tract the square root of the product , and the result will be the dis- 
tance in miles. 

I. At what distance at sea can Mt. Aconcagua be seen, if its 
height is known to be 24000 feet? 

By formula, c=V§a~=V%X 24000=^36000=1 90 mi., nearly. 
{ 1. 24000 ft.=the height of the mountain 
. II.<{2. 1X2 4000= 36000. 

[3. .-. V36OO0=10V360=190 mi., nearly. 
III. .-. Mt. Aconcagua can be seen at a distance of 190 miles. 

Prob. CJLXXI, Given the sum of the hypotenuse and 
perpendicular, and the base, to find the perpendicular. 

Formula. — p= ^ — , where s is the sum of the * 

ZiS 

potenuse and perpendicular, and b the base. 



MENSURATION. 289 

Rule. — 1- From the square of the sum of the hypotenuse 
and perpendicular subtract the square of the base, and divide the 
difference by twice the sum of the hypotenuse and perpendicular. 

2. To find the hypotenuse : To the square of the sum of the 
hypotenuse and perpendicular, add the square of the base and di- 
vide this sum by twice the sum of the hypotenuse and perpeiidic- 
ular. 

1. A tree 120 feet high is broken off but not severed. The 
top strikes the ground 34 feet from the foot of the tree; what is 
the height of the stump? 

s ?—b 2 120 2 — 34 2 [stump. 

By formula, p= — = 9 >Q =55|fr ft, the height of the 

1. 120 ft.=the sum of the hypotenuse and perpendicular. 

2. 34 ft.=the base, or the distance the top strikes from the 
foot of th : tree. 

JI.<|3. 14400 sq. ft.=120 2 =the square of said sum, 

4. 1156 sq. ft=34 2 =the square of the base, and 

5. 14400 sq. ft.— 1156 sq. ft=13244 sq. ft=the difference. 

6. .-. 13244-:-(2xl20)=55ii ft.=the height of the stump. 

III. .'. The height of the stump is 55-g^ feet. 

Note. — This rule is easily derived from an algebraic solution. Thus: 

Let x=the perpendicular,^ — .v=the hypotenuse, and £=the base. Then, 

s-—b 2 
x 2Ji t b 2 =z{s—x) 2 , or .v 2 -f^ 2 =.? 2 — 2-v-v+x 2 , and v— — . 

Prob. CLXXIT. To find at what distance from the large 
end of the frustum of a right pyramid, a plane must be passed 
parallel to the base so that the two parts shall have 
equal solidities. 

3 V 

Formula. — /*=- . , where V is the 

volume of the frustum, B the area of the lower base, B 2 the area 
of the "dividing base," and 4 . I>B 2 the area of the mean base be- 
tween the "dividing base" and and lower base. 

Rule. — 1. Find the volume of the frustum by Prob. XCIII. 

2. Find the dimensions of the "dividing base" hy extracting- the 
cube root of half the sum of the cubes of the homologous di?nensions 
of the upper and lower bases. Then find the area of the ""divid- 
ing base." 

3. Divide half the volume of the frustum by one-third of the 
sum of the areas of the lower base, "dividing base," and mean 
base between them, and the quotient will be the length of the 
lower part. 

I. How far from the large end must a stick of timber, 20 feet 
long, 5 inches square at one end and 10 inches square at the 
other, be sawed in two parts, to have equal solidities? 



290 



FINKEL'S SOLUTION BOOK. 



By formula, h— 



sv 



2(Jt-H(B#*)+B 2 ) 



3xM* 2 +&4* 2 ) 



240(10 2 +10x5+5 2 ) 



+oryi 2p , +«xvK^)+ii](^)r] 



42000 



1680 



2(100+25^36+-^ / 6) 8+2^36+^6 
_ 1680 1680 7fim ,. 

8+6.603855+5.4513618 20.0552168 < uoo °-r • 

Construction. — Let ^4i> CD — E be the piece of timber, AB CD 
the lower base, EFGH the upper base, and OL the altitude. 
Prolong the edges AH, BE, CF, and DG and the altitude OL 
till they meet in P. Draw KL to the middle point of AD, OI 
to the middle point of GH and draw PIK. Let SMNR be 
the dividing base. 

f 1. AB=10 in.=£, the side of the lower 
base. 

2. HE=5 \n.=c, the side of the upper 
base, and 

3. OL=20 ft.=240 in.=a-, the altitude. 

4. A;G=AZ-eZ(=/O)=-i(3-c)=4(10 
in. — 5 in)=2^in. By similar triangles, 

5. KQ: QJ..KL.PL, or \(b—c)\av^PL. 
Whence, 

PL=^L=±0 ft 



II. 



6. 



7. 



6-c 
PO=PL 



OL-- 



ab 



b—c 



ac 




=20 ft. 

8. z>=|/>0X^ 2 =i^ 2 =4X24Ox5 2 

=2000 cu. in., the volume of the pyramid HEFG. 

9. V=iOZx(AB 2 -{-ABxL/E+//E 2 )=^a(b 2 +bc 

+c 2 )=14000 cu. in., the volume of the frustum 
ABCD—E. 

10. .-. \ V=i of 14000 cu. in.=7000cu. in., the volume of 

each part. 

11. ^+|7/=2000 cu, in.+7000 cu. in.=9000 cu. in., the 

volume of the pyramid, SMNR — P, and 

12. z;+ F=2000 cu. in. +14000 cu. in.=16000 cu. in., the 

volume of the pyramid ABCD — P. By the princi- 
ple of similar solids, \AB Z , or 

13. HEFG-P : SMNP-P : ABCD-P : : HE* : SM* : 

14. v : v+i V: v-\- V. ; c 3 : SM* : bK But 



MENSURATION 291 

15. z,_(_| V=i[v+(v+ V)], I e., v+i V, or SMNRPis an 
arithmetical mean between v and v-\- V, or HEFG 
— P and AB CD— P. 

17. .'. SM s =%(c*+& B ),l e, SM* is an arithmetical mean 
between J^i? 3 and AB*, or c 3 and £ 3 . Whence, 

18. SM^y^c? +6* )]==y[±(5* +10* )]=§&' 36= 
8.2548188+in. 

19. SM 2 ==$/[i(c'*+?>2)¥=(iySQy==¥yQ=68.U202 
sq. in=the area of the dividing base. 

20 sJ(SM*xA£ 2 )=SMxA£=iySQx 10=25^36= 

82.54818 sq. in.=the area of the mean base of the 
part cut from the frustum. 

21. .-. iLT(AB' i +SMxAB+SM <i )=iLT{b' 1 

+1^(36) Xl0+(4y36) 2 ]=ii/r(100+82.54818 
+68.14202 )= \L TX 250.6902 = ZTx 83.5634 = the 
volume of the frustum A BCZ>—M .But \—M. 

23. \ F=7000 cu. in.=tbe volume of the frustum ABCD 

24. ••• LTX 83.5(534=7000 cu. in. Whence, 

25. L 7=7000^-83.5634=83.76883 in.=6 ft. 11.76883 in., 
the length. 

III. /. The stick must be cut in two at a distance of 83-76883 
in., or 6 ft. 11.76883 in., from the large end. 

N OTE . — The frustum of a cone may be divided into two equal parts in 
the same manner. The frustum of a pyramid or a cone can be divided into 
any number of equal parts on the same principle as that for dividing a 
trapezoid into;/ equal parts, Prob. CLXI. 

I. The area of a rectangle whose length is 20 rods is 120 sq. 
rods ; what is the area of a similar rectangle whose length is 30 
rods? 

Principle, — Similar areas arc to each other as the squares 
of their like di mentions or as the sqtiares of any other ko?7iologous 
lines, 

1. 30 rods=the length of the given rectangle, and 

2. 120 sq. rd.=its area. 
II. -J 3. 30 rods=the length of the required rectangle. 

|4. .-. 20 2 :30 2 ::120sq. rd. : (?). Whence, 
[5. ?=(120x30 2 )-^20 2 =270 sq. rd. 
III. .-. The area of the rectangle is 270 sq. rd. 

I. The area of a rectangle whose width is 7 feet, is 210 sq. ft. ; 
what is the length of a similar rectangle whose area is2100sq. ft. 

1. 210 sq. ft.=the area of given rectangle, and 

2. 7 ft.=its width. Then 

TT 3. 210-f-7=30 ft.=its length. 

"14. ••• 210 sq. ft: 2100 sq. ft. : : 30 2 :(?). Whence, 

5. ?=( 21 00 X30 2 )-^2 10=300 ft.=the length of the re- 
[ quired rectangle. 



292 



FINKEL'S SOLUTION BOOK. 



II. 



III. 



III. The length of the required rectangle is 300 feet. 

I. If the weight of a well proportioned man, 5 feet in height, 
be 125 lbs., what will be the weight of a similarly proportioned 
man 6 feet high ? 

Pvifldple, — Similar solids are to each other as the cubes of 
their like dimensions or as the cubes of any other homologous lines. 

1. 5 ft.=the height of the first man, and 

2. 125 lbs.=his weight. 

3. 6 ft.=the height of the second man. 

4. .-. 5 3 : 6 3 : : 125 lbs : ( ?). Whence, 

5. ?=(125X6 3 )-4-5 3 =216 lbs., the weight of the second 
man. 

•.The weight of the man whose height is 6 feet, is 216 lbs. 

I. James Page has a circular garden 10 rods in diameter. How 
many trees can be set in it so that no two shall be within 10 feet 
of each other and no tree within 2-J- feet of the fence? 

Constntction. — ~LetAB C be the circular garden, A C it diameter, 
and O its, center. Then with Oasa center and radius AO = -Jot 
(10X16-J- ft.— 2X2| ft), or 80 ft, discribe the circle abedef, 
and in it describe the regular hexagon abedef. Then aO 
—ab=&) ft. Begin at the center of the circle and put 
8 trees 10 ft. apart on each radii, aO, bO, cO, dO. eO, and fO. 
Then joining these points by lines drawn parallel to the diame- 
ter of the circle as shown in the 
figure, their points of intersec- 
tion will mark the position of the 
trees. Hence, the trees are ar- 
ranged in hexagonal form about 
the center. The first hexagonal 
row contains 6 trees, the second, 
12, the third 18, and so on. Since 
the radius of the circle on which 
the trees are placed is 80 feet and 
the trees 10 feet apart, there will 
be 8 hexagonal rows. 

1. 6=the number of trees in the first hexagonal row. 

2. 12=the number of trees in the second hexagonal row. 

3. 48=the number of trees in the eighth hexagonal row. 

4. .-. 216=|(6+48) X8=the number of trees in the eight 
hexagonal rows. 

5. 24=6x4=the number of trees at the sides of the hexa- 
gon abedef. 

6. .-. 216+24+1, the tree at the center ,=241=the number 
of trees that can be set in the garden. 




II. 



MENSURATION. 



293 



UA 



the 



of 



III. .*. There can be set in the garden, 241 trees. 

( Greenleaf's JVafl Arith., f. £44, firob. 25.) 

I. There is a ball 12 feet in diameter on top of a pole 60 feet 
high. On the ball stands a man whose eye is 6 feet above the 
ball; how much ground beneath the ball is invisible to him? 

Construction. — Let BF be the pole, JL the center of ball, and 
A the position of the man's eye. Draw AFC tangent to the ball 
at F and draw BF and BC. Then the triangle AFL is right- 
angled at F. 

(1. 60 ft.=BF, the length of 
pole. 

2. 12 ft.=FD, the diameter 
the ball, and 

3. 6 ft.=AB>, the height of the 
man's eye above the ball. 

4. 12 it.=AD-\-DL=AL. Now 

5. AF=sj(AL*—LFz) 
=V(12 2 — 6 2 )=6y/3ft. In the 
similar triangles ALF and 

ACB, 
AF:LF::AB: BC, or 
6 V /3 ft. :6 ft. ::(6ft.+12ft. 
+60 ft.), or 78 ft.: BC. 
BC=(6 x78)-f-6V3=78 
■^-V3=ix78V3==26V3 ft. 
ttBC* = it (26^3)2=6371.14989 



7. .-. 




sq. 



FIG. 68. 
ft.=the area 



the circle over which the man can not see. 



of 



III. .-. 6371.1498932 sq. ft.=the area of the invisible ground 
beneath the ball. 

I. Three women own a ball of yarn 4 inches in diameter. 
How much of the diameter of the ball must each wind off, so that 
the may share equally? 

1. 4 in.=the diameter of the ball. 



II. 



9. 
10. 



Then 

l7r(4)J= 3 ^^=the volume of the ball. 
i of \ 2 7T== , V 7r == eac h woman's share. 
¥ 7r —V 7r=r, / 7r = tne volume of the ball after the first 

has unwound her share. But 
■J-7rZ> 3 =the volume of any sphere whose diameter is D 
.\\nD*=^n. Whence, 
Z>3 =V 7r -H 7r = 1 | 8 ,and 

Z>==y i i8=4^|=|^18=4x2.6207414=3.4943219 in., 
diameter of the ball after the first unwound her share. 

.-. 4 in.— 3.4943219 in=.5056781 in., what the diameter 
was reduced by the first woman. 

"¥-*— V^— ¥*> the volume of the ball after the 

second had unwound her share. 



294 



FINKEL'S SOLUTION BOOK. 



12. 



III. 



■V ,7! 
I 2 -' 



11. .-. y( 3 9 2 ^-H^)=4y^|^9=|X2.0800837 

=2.5734448 in., the diameter of the ball after the sec- 
ond woman unwoud her share. 
.-. 3.4943219 in.— 2.5734448 in.=.7208771 in., what *he 
diameter was reduced by the second woman. 

The diameter was diminished .5056781 in. by the first 

woman, 
.7208771 in. by the second woman, and 
.7734448 in. by the third woman. 

{Milne's Prac. Arith.,f. 335, prod. 8.) 

Note.— The following are the formulas to divide a sphere into n equal 
parts, the parts being concentric: 

-«J(^)]-^.--.=[-JC^)-*^)]^ 

is the diameter of the sphere; Z> 15 the diameter after the first 
part is taken off; D 2 , the diameter after the second part is taken 
off; and soon. Then D—D x , D 1 —D 2 , &c, are portions of 
the diameter taken off by each part. 

I. A park 20 rods square is surrounded by a drive which con- 
tains T \ T ^ of the whole park; what is the ■■^■nnnnpi 
width of the drive ? 

1. 20 rd.=AD=DC, a side of the 
park. 

2. 400 sq. rd.=20 2 =the area of the 
park AB CD. 

3. T V 9 o of 400 sq. rd.=76 sq. rd.=the 
area of the path. 

4. 400 sq. rd.— 76 sq. rd-=324 sq. rd. 
=the area of the square EFGH. . FIG. 69. 

5. EF=sJ (Z24)=18 rd., the side of the square EFGH. 

6. .'. IH— EF=20rd.— 18rd.=2 rd., twice the width of the 
path. 

.7. .-. 1 rd.=| of 2 rd.=the width of the path. 
III. .*. The width of the path is 1 rod. 

I. My lot contains 135 sq. rd., and the breadth is to the length 
as 3 to 5 ; what is the width of a road which shall extend from 
one corner half around the lot and occupy \ of the ground. 

Construction. —luzK. AB CD be the lot, and DABSNR the 
road. Produce AB, till BE is equal to AD. Then AE is 
equal to AB-\-AD. On AE, construct the square AEFG, and 



II. 




II. 




MENSURATION. 295 

on J? J? and GF respectively, lay off EI and FK equal to AB. 
Then construct the rectangles BEIH, ILKF, and KMDG. 
They will each be equal to A B CD, for their lengths and widths 
are equal to the length and width of AB CD. Continue the road 
around the square. Then the area of the road around the square is 
four times the area of the road DABSNR. 

( I. J=thc width AD of the 

lot. Then 

2. §=the length AB. 

3. fxf=135 sq. rd., the 
area of the lot. 

4. £Xf=4 of 135 sq. rd. 
=27 sq. rd., and 

5- |X|=(4) 2 =3 times 27 
sq.rd.=81 sq. rd. 

6. .-. |==y81=9 rd.,the 
width AD, 

7. J=J of 9 rd.=3 rd., and 

8. f=5 times 3 rd.=15 
rd.,the length AB. 

9. 15 rd+9 rd.=24 rd.= FIG. 
AE, the side of the square AJ.FG. 

10. .*. 576 sq. rd.=24 2 = the area of the square AEFG. 

11. 33f sq. rd.=^ of 135 sq. rd.=the area of the road 
DABSNR. 

12. .'. 135 sq. rd.=4x33j sq. rd.=the area of the road 
around the square. Then 

13. 576 sq. rd. — 135 sq. rd.=441 sq. rd., the area of the 
square NOPQ. 

14. .-. 21rd.=V441 =NO, a side of the square NOPQ. 

15. AE— NO=24 rd.— 21 rd.=3rd.=twice the width of 
the road. 

16. .-. H rd.=24j ft.=-J of 3 rd.=the width of the road. 

III. .-. The width of the road is 24j ft. 

(B. H. A.,J>.J,07,prob.99.) 

I. The length and breadth of a ceiling are as 6 to 5 ; if each 
dimension were one foot longer, the area would be 304 sq. ft. ; 
what are the dimensions? 

Co7istructio)i. — Let ABCD be the ceiling, AB its width and 
BC its length. Let A/GEbe the ceiling when each dimension 
is increased one foot. On BC, lay off ^A'equal to AB and draw 
LK parallel to AB. Then ABKL is a square whose side is 
the width of the ceiling. 






296 



FINKEL'S SOLUTION BOOK. 



1. i=A£, the width of the ceil- 

ing. "Then 

2. %=BC, the length, and 

3- tXf=-4^X^C=thearea6f 

the ceiling. 

4. iXl=£CxBA BI being 1 

foot,=the area of the rect- 
angle BCHI. 

5. iXl=£>CxCJ?, CF being 

1 foot,=the area of the rect- 




10 
11 
12 
13 



angle DCFE. 
1 sq. ft.=l 2 =the area of the 
square CFGH. 
•••fXf+tXl+fXl + 1 sq.ft. 
=the area of AIGE. But nG 7 

fXl+fXl=VXl=rectangle^/^C+rectangle 
DCFE, i. e.,it equals a rectangle whose length is 
§+h or V , and width 1 ft. 

••• fXi+i X }+\X\+^- ft=!Xf+Vxl+l sq. ft. 
=the area of AIGE. But 
304 sq. ft.=the area of AIGE. 

6 Xf+y Xl+1 sq. ft.=304sq. ft. Whence 



; |-X|+VXl=303 sq.ft==the area of AIHCFE. But 

is y ft. ; for a rectangle whose 



1 1 1 v 5 



in w 



hich 



length is y , and the width 1 ft, has the same area . 
a rectangle whose width is y ft. and length 1, or 4. 
•'• f Xf+y Xf=303sq. ft., in which U is V ft. 

W + H^ f=50i sq ' a== * of (tx*+Vxf)=i of 

6v6 sq. tt., 

* Xf+y Xf=252i sq. ft.=5 X (iX|+iiX!)=5x50i 
sq. ft. But 

f Xf=the area of the square ABKL, and 
V Xf=the area of the rectangle ALNP whose length 
AL is I and width LN y ft. 

iof(VXf)=HXf=half the rectangle ALNP= the 
rectangle OMNP, which put to the side AB of the 
square ABKL as in the figure. 

f Xf+y Xf=252| sq. ft.=the area of SRA OMK. 
\\\ sq. ft.=(|J)2 = theareaof the square RQOA, since 
AR is 14 ft 

•••fX|+Vxf+Hi sq. ft=252| sq. ft+i|l sq. ft. 
== a ftt i s q- ft.=the area of ( 67?^ J^4-i? <X4 ). 
=the area of SQMK. 

f+H ft.=^f|} i =W ft.=the side SKoi the square 
SQMK. 
-Ly ft— J i ft.=y a o ft.=15 ft=67T— SB=BK 



14. 
15. 

16. 

17. 

18. 

19. 



20. 
21. 

22. 



23. 

24. 



=AB y the width of the ceiling. 



MENSURATION, 



297 



25. i- 
126. * 



III. 



=f of 15 ft.=3 ft., and 

=6 times 3 it. =18 ft =75 C, the length of the ceiling. 
^ 15 ft.=the width of the ceiling, and 
I 18 ft.=the length. 
Remark. — In this solution there is but one algebraic operation ; 
viz., extracting the square root of the binomial expression, 
(4X!_r-i^ Xf+xii sc b * n -)» * n ste P 23. This might have been 
omitted and then the solution would have been purely arithmeti- 
cal ; for, the area of the square SQMK being known, as shown 
by step 22, its side iS-AT could have been found by simplv extract- 
ing the square root of its area, 8 -}-!!- 1 SC L- ^- Then by subtracting 
SB, which is \\ ft, from SK, we would get BK{=AB), the 
width of the ceiling. 

The following solution is quite often given in the schoolroom: 
304-r- (5X6) = 10+. V10 = 3+. 

5 x3= 15, the width and G X 3=18, the length. 

I. A tin vessel, having a circular mouth 9 inches in diameter, 
a bottom A\ inches in diameter, and a depth of 10 inches, is £ part 
full of water ; what is the dia.r.eter of a ball which can be put 
in and just be covered by the water? 

ConstructioJi. — Let ABCD be a vertical section of the vessel, 
AB the top diameter, DC the bottom diameter, and EF the al- 
titude. Produce AD, BC, and EF till they meet in G. Draw 
MC parallel to EF. In the triangle A CB inscribe the largest 
circle IEP and let Q be its center. Draw the radius IQ. Now 
( 1. A£=iAB=I?—4i in.=the radius of the mouth. 

2. CF—%I?C—r=2£ in., the radius of the bottom, and 

3. EF—a=\Q in., the altitude of the vessel. 

4. M£=EB— BM(=FC)=R—r=A\m. 
— 2±in.=2| in. 
BMC and BEG, 

MB.MC: -EB\EG, or 
R—r :a::R:EG. Whence, 
aR _ 10X4^ 
7?— r 4^— 2f 
of the triangle A GB. 

2AAGB ABxEG 



5. 



In the similar triangles 



6. EG: 



; =20 



the altitude 



7. /0= 



2Ra 



AB+AG+BG 
But 



8. 




AB+BG+BG 

AG=BG==<(EBi+EGi)= s [R~ + (2ay]:= s >\(UP 
+20 2 ]=V(420±)=20i in. 

4/fe _iiX20_o3 

s+2o;v _: ' 



9. /. /<? 



in., the radius 



2i?+2V(i? 2 +« 2 ) 
of the largest sphere that can be put in the vessel or in 



298 



FINKEL'S SOLUTION BOOK. 



II. 



10 



12 



13 



14. 



16. 



17 



18 



19. 
20. 



the cone A GB. 



n 



Rla 



) 3 =l^(3 



VLL3 n 
125 a 



=the volume of the largest sphere that can be put 
in the cone AGB. 
11. i EGxnEB*=±7t2aR*=±7t x20X(41) 2 =135tt, the 
volume of the cone A GB. 

,. i^a^-i < ^. +v( ^f +2a2) ) S -frrto*' X 

E9/y2 /p — , 
1 I^TT 7776-* 9099-. 

the quantity of water in the cone which will just cover 
the largest ball that can be put in the cone A GB. 
i7tBGX^C 2 =i7rar^==^7rXl0xm) 2 =H^, the 

volume of the cone DGC. 
.-. ^nar 2 -\-\ of the volume of the vessel = i-^7t-\-^ of 
the volume of the vessel— the quantity of water in 
the cone necessary to cover the required ball. But 
15. \na(R*+Rr+r* )=i7rl0[(4i) 2 +4|X2i+(2i)2] 
-af^tf, the volume of the vessel, by Prob. XCIII. 

^n ar 2 -\-\ of the volume of the vessel ==^nar 2 -\-^ of 
i7ta(R 2 +Rr+r 2 )=±7ta[r 2 -{-i(R 2 +Rr-\-r 2 )]=^ 
rt-\-\ of 2-^$-7r=^^7T , the quantity necessary to cover 
the required ball. 

The quantity of water necessary to cover the largest 
ball: the quantity of water necessary to cover the 
required ball : : (radius) 3 of largest ball : (radius) 3 of 
required ball. Hence, 



[i?+V(i? 2 +4« 2 )] 



9_0_9_9 
1 2 5 



:^(337) : 



21. HO- 



4* 



■■^(V) 



HO' 

+>J(B 2 +4a- ' 

*\s - HOK Whence, 
HO. Whence, 

2 +K^ 2 +^+^ 2 )] 

2a*R 



or 



7r:i||5-7r::(3f) 3 :HOK 



").=H^( ¥)X3|] 



22. 



[R + ^R* + 4a 2 )J 

^-|^337=9^(6 5 T \), and 

^ (6T ¥ )=6.1967+in., the diameter of the required 

ball. 

III. .'. The diameter of the required ball is 6.1967+ in. 

I. I have a garden in the form of an equilateral triangle 
whose sides are 200 feet. At each corner stands a tower; the 
height of the first tower is 30 feet, the second 40 feet, and the 
third 50 feet. At what distance from the base of each tower 



MENSURATION 



299 



must a ladder be placed, so that without moving it at the base it 
may just reach the top of each, and what is the length of the 
ladder? 



Construction. — Let ABC be the triangular garden and AD, 
BE, and CE the towers at the corner-. Connect the tops of the 
towers by the lines ED and DE. 
From G and H, the middle points 
of DE and DE, draw GMand HN 
perpendicular to DE and DE, and 
at Mand N draw pcrpendicularsto 
AB and A C in the triangle ABC, 
meeting at 0. Then is equally 
distant from D and E. For, since 
M is equally distant from D and E, 
and MO perpendicular to the plane 
ABED, every point of MO is 
equally distant from D and E. For 
a like reason, every point of NO is 
equally distant from D and E ; hence, O their point of intersec- 
tion, is equally distant from D, E, and E and is, therefore, the 
point where the ladder must be placed. Draw Df and DJ par- 
allel to AB and A C, GA^and HL perpendicular to AB and A C, 
MP perpendicular to A C, and OR parallel to NP. Draw the 
lines OB, OC, and OA, the required distances from the base of 
the ladder to the bases of the towers. Draw EO, the length of 
the ladder. 




8. 
9. 

10. 



AB=B C=A C=200 ft.==j, the side of the triangle. 
EC=bO ft. =a, the height of the first tower, 
EB=40 ft.=£, the height of the second tower, and 
AD=Z0 ft.=c, the height of the third tower. Let 
h =^[AB>-(iA C) 2 ]=V[j 2 — (^)2]=4V3j=100 

^3 ft=the perpendicular from B to the side A C. 
E7=BE—B/(=AD)=(d—c)=40 ft— 30 ft. 

=10 ft. 
GK=i(££+AZ))=i(&+c)=i(40 ft.-f30 ft.) 

=35 ft. In the similar triangles DIE and GKM, 
DIJE : : GK: KM, or s : £— c : : 4-(M- c ) : KM. 
. KM ^-c 2 _40*-30 2 



=1011 ft., and 



**4 



^2_ C 2^ 5 2_|_^2_ ( 



2* 



! 4-c 2 — 3 2 



c2 I £2 c 2 

25 2s 

=98^ ft. In like manner, 
12. JfL=i(a+c)=i(bO ft.+30 ft.)=40 ft., 



300 



FINKEL'S SOLUTION BOOK. 



r A ^ 



II. 



KB. 



13. ZiV= 



2* 



~=4 ft., 



14. AJV=A/.+ZN=$s-{ 



a*—c 2 s 2 -\-a' 



2s 
s 2 -\-a 2 — c M 



2s 

s 2 -\-c 2 — a 2 



104 ft. 



15. JVC=AC— AN^ 

2s 2s 

96 ft. By similar triangles, 

16. AB:AZ :: AM: AP, or s:\s : ( 5 2 +£ 2 — c~ ) -f- 2.? 

\AP. Whence, 

17. AP=(s 2 -\-b 2 — c 2 )-M*=50i ft- 

18- .\PL=AL— AP=[±s— (s 2 -\-h 2 — C 2)^_4 5 ] = 
( 5 2_|_£2__£2 )^4 5==: 49i. ft. 

19. i?0=PiV=PZ+Z^Vp=(5 2 +c 2 — b 2 )~-4s + (a 2 

— c *)^-2s=(s 2 +2a 2 —b 2 —c 2 )-^As=bZ\ ft. By 
similar triangles, 

20. AB : BZ: :AM: MP, or s: ±>J3s::(s 2 +b 2 — c 2 )^-2s: 

MP. Whence, [lar triangles, 

21. j\fP=[(s 2 +b 2 — c 2 )-^45]X\/3=oO|V3ft. Bysimi- 

22. MP :AP ::RO:RM, or [ ( s 2 -\-b 2 — c 2 ) -t-4?]\/3 : 

( 5 2_|_32__ c 2 )_^_4 5 . . ( S 2_|_ 2 a 2 — ^2— C 2 )-2-4j : 7?JZ. 

23. RM=(s?+2a 2 —b 2 —c 2 )4<JZs=[(s 2 +2a 2 —b 2 — 

c 2 )-=-12j]V3=17HV3 ft Again 

24. MP: MA ::RO: OM, or [(j2-|-J 2 — c 8 )-^4s]V3 : 

(5 2 +^2_ c 2^_ h 25: : (s 2 +2a%— b 2 — c 2 )^-As : OM. 

25. .-. OM=(s 2 +2a 2 —b 2 —c 2 )-i-2s/Ss=[(s 2 -\-2a 2 —b 2 

—c 2 )H-6^]V3=35AV3 ft. 

26. 7V=i? P=MP—R M=\(s 2 +b 2 —c 2 )h-4s]V3— 
( * 2 + 2« 2 — /S 2 — c 2 )^-12j]^3=[ (s*—a 2 -\-2b 2 — c 2 ) 

-^-65]V3=33iV3ft. Then 

27. OCW( OJV 2 + JVC 2 )=\[ r 2 - ^+ 2 ^~ g2 V3) 2 

+ 2j j ]=V[(33iV3) 2 +96^]=V12516 1 J, 
=111.8796+ft. 

28. OA=^( ON 2 +AN 2 )=J [(^Z^^^X 

+(^ 
=V14116 T V=118.81il+ft. 

29. OZ?=V( Q^+^g» )= Jr C' +2a ' fi 7^ , ~ Cl V 3 1 
& J ]=V[(35 T \V3) 2 +(98i) 2 J 



) 2 ]=V[(33iV3) 2 +104^] 



2s 

=iV214657i=H5.8278+ft. 
1. OZ , =V(-^^ 2 +^^ 2 )=\ / [(i\ / 214657i) 2 +40 2 ], 
=V(13416|+1600)=V15016 T 1 2=122.5402+ft.=the 
length of the ladder. 



MENSURATION. 



301 



ni..-.<! 



1. 111.8796-f-ft.=the distance from base of the ladder- 
to the base of the tower FC, 

2. 1 18.81 ll+ft.=the distance from the base of the lad- 
der to the bast of the tower AD. 

3. 115.8278+ft.=the distance from the base of the lad- 
der to the bsse of the tower BE, and 

4. 122.5402+ft.=the length of the ladder. 

( Greenleafs Nai'l Arith., p. J^JfJf, prob. 38.) 
Remark. — When the sides of the triangle are unequal, pro- 
ceed in the same manner as above. In some cases the base of 
the ladder will fail without the triangle. 

I. At the extremities of the diameter of a circular garden 
stands two trees, one 20 feet high and the other 30 feet high. At 
what point on the circumference must a ladder be placed so that 
without moving it at the base it will reach to the top of each tree, 
the diameter of the garden being 40 feet. 

Construction. — Let ABC be the circular garden and A C its 
diameter, ; nd let AB and CD be the two trees at the extremi- 
ties of the diameter. Connect the tops of the trees by the line 
FD and from the middle point B of BD let fall the the per- 
pendicular BH. Draw EG perpendicular to BD. Then all 
points of EG are equally distant from BD. At G draw BG 
perpendicular to AC. Then all points of BG are equally dis- 
tant from B and D. Hence, B is the required point. 

1. A C=2/i=40ft., the diameter of the garden. 

2. CD=a=?>0 ft., the height of the tree CD, and 

3. AF—.b=3& ft., the height of the tree AF. 

4. DI=DC—CJ(--=AF) = a — b 
=-40 ft.— 30 ft.=l0 ft. 

5. EH--= i ( CDA- AB) = i( a+b) 
=4(40 ft.+30 ft.) =35 ft. By 
similar triangles, 

6. BB.ID-.EH.HG, or 
2B:a—b: : \(a+b) : HG 
Whence, 

/,2 A2 

7. J/G=^-r^-=SZ f. 



II. 




FIG. 74. 



[B 



8. GB^(BW-HG*)=^[R*-(^^y~\= 

V[i6/? 4 — (a 2 — b 2 ) 



4B 



= Vn'23 ft. 



9. 



AB=>J(AG 2 +GB 2 )=\[(AR+JBG) 2 +(GB 2 )]- 
2 K)/? 4 — (a°~— b 2 ) 2 - 



J[0^?) 



16/?* 



] 



=5V46ft. 



=34.91165 ft., nearly, and 



M> -V8C 



AC 



^ V ^ 












V* 






c^ 



* 



. b'V* 



v— * 



302 



FINKEL'S SOLUTION BOOK. 



10. £C=>1(GC*+GB*) 

16P+—(a~—b 2 ) 2 



~| =f V82 ft.=ll. 31942 ft. 



III. 



I 16i? 2 

34.91165 ft. the distance from the smaller tree, and 
11.31942 ft. the distance from the larger tree. 

I. Seven men bcught a grindstone 5 feet in diameter ; what 
part of the diameter must each grind off so that they may share 
equally? 

Construction. — Let AB be the diameter of the grind stone, 
O its center, and A O its radius. From A draw any indefinite 
line ^47Vand on it lay off any convenient unit of length seven 
times, beginning at A. LetT 5 
be the last point of division. 
Draw OP ', and from the other 
points of division draw lines 
parallel to OP, in intersect- 
ing the radius A O, in the 
points /, e, d, c, b, and a. Then 
the radius is divided into 
seven equal parts. On radius 
AO, as a diameter describe a 
semi-circumference A 0-AT,and 
at a, b, c, d, e, and/*, erect per- 
pendiculars intersecting the 
semi-circumference in M, £, 

K, I, H, and G. Then with /r/Q § ^5, 

O as a center and radii equal the chords MO, LO, KO, IO, 
HO, and GO, describe the circles as shown in the figure. Then 
each man's share will be the area lying between the circumfer- 
ences of these circles. For, the chord G0 2 =Gf 2 -\-/0 2 and, by 
a property of the circle, Gf 2 =AJXfO. .-. G0 2 =AfxfO+f0 2 , 
== (Af-\-fO)fO=AOxfO=} t A O 2 . In like manner H0 2 =A O 
Xe0=\AO 2 , X0 2 =:AOxdO-- 




\A0 2 , &c. 



A B=D=5 ft., the diameter of the grind stone. 

AO=jR=2$ ft, the radius. 

.-. 7ri? 2 =7rX(2|) 2 =6^zr— the area of the stone. 



4. \ of 7rR 2 =\7tR 2 =\ of 6^7r=||-7r=each man's share. 



5. Utt- 



-|-|7r=^7r=the area of the stone after the first has 



ground off his share. 

6. .•. A /(if7T~7r)= 1 5 ¥ v'42=2.314554-ft., the radius MO. 

7. 2( AG— MO)=2(2^ i't.—j 5 ^42 ft.) =2(2£ ft.— 2.31455 

ft.)=.3709 ft., part of the diameter the first grinds off. 

8. 6^7r — 2 of6^7T= 1 2 2 ¥ 5 7r=the area after the second grinds 

off his share. 



MENSURATION. 303 

9. .-. V(W 7r ^~ 7t )=Wj =2.112875 ft., the radius ZO. 
Then 

10. 2(MO— LO)=2(qV t *t— Wy )=2(2.31455 ft.— 
2.112875 ft.)=.40335 ft, the part of the diameter the 
second grinds off. 

11. iS\n — f cf 6^7T= 2 y 5 7T=thc area after the third has 
ground off his share. 

12. .-. V( 3 T 57r -^ 7r )=fVT= :5 V /1 -=l-889822+ ft, the radius 
KO. Then, 

13. 2(ZO— A"0)=2(|Vf— 5Vt)=2(2.112875 ft— 
1.8899822 rt)=.44610G ft, the part of the diameter 

II. <j the third grinds off. 

14. 6j7T — 4. f gi. 7r -_T.|.- r __ t i ie area a fter the fourth has 
ground off his his share. 

15. .'. \/(B 7r -^- 7r )=-iVf=l-636634 ft., the radius7(A Then 

16. 2(KO— /0)=-2(5V4--M)=-2(l 889822 ft.— 
1.636634 ft.)=.506S76 ft., the part of the diameter the 
fourth grinds oft'. 

17. 6^7T — f of Q\7T=l^ 7r=the area after the fifth grinds 
off his share. 

18. .-. V(i8 7r -^ 7r )= ; }V / T= 1 - 336306 ft -> the radius HO. 
Then 

19. 2(10— HO)=2(^l— #\/f )=2(1.636634 ft- 
1.336306 ft)=.600656^t.,' the part of the diameter the 

fifth grinds off. 

20. 61tT — -f of 6;j7r=!f 7r=the area after the sixth grinds 
off his share. 

21. .'. V(2f7r- r -7r)=-:H / i-=.949911 ft., theradius GO. Then 

22. 2(1/0— GO)=2(? z Vt—isT)— 2 ( 1-336306 ft— 
.944911 ft.)=.782790 ft, the part of the diameter the 
sixth grinds off". 

23. 2X-944911 ft— 1. 889822 ft., the diameter of the part 
belonging to the seventh man. 

I. J. A. M., having a wooden ball 2 feet in diameter, bored a 
hole 1 loot in diamer through the center. What is the volume 
bored out? 

Construction. — Let ABCDEF be a great circle of the ball 
and let A CDF be a verticle section of the 
auger hole. Draw the diameter BOE 
and the radius AG. Then the volume 
bored out consists of a cylinder, of which 
ACDF is a vertical section, and two 
spherical segments, of which ACB and 
FDE are verticle sections. 

(1. BE=2 feet=2i?, the radius of 

the ball, and 
2. A 0=1 foot=2r, the radius of the 

auger hole. fig. 76 




304 



FINKEL'S SOLUTION BOOK. 



3. iAJ?=0G=iJ(A0 2 — AG* )=</(!?*— r*)=ys. 

4. .". AF=2\(B 2 — r) 2 =V3, the length of the cylinder. 
II. \ 5. .-. V—nr 2 X(V3)=i 7r V3» the volnmeof the cylinder, and 

, 2V / =2(iBGxxAG 2 +±7r£G*)=[J?— s/(P 2 — r*)] 
x 7tr 2 +in:[jR— V(X 2 — r 2 )] 3 
i7r(l— |V3)+i7r(l— iV3) 3 = T V^(16— 9^3), the vol- 
ume ot the two spherical segments. 

, .-. F+2F / =i7r v / 3+ T 1 2 7r(16— 9V3)=i7r(8— 3V3), 
=1.46809 cu. ft.==2536.85952cu.in. 

III. .-. The volume bored out is 2536.85952 cu. in. 



I. What is the diameter of the largest circular ring that can 
be put in a cubical box whose edge is 1 foot? 

Construction. — LetABCD — E be the cubical box. I et 7, K, 
L, M, N, and P, be the middle points of the edge CF, GF, GH, 
HA, AB , and BC respec- 
tively. Connect these points 
by the lines AY, XL, LH, 
MN, NP, and PI. Then 
IKLMNP is a regular 
hexagon, and the largest 
ring that can be put in the 
box will be the inscribed cri- 
cle of the hexagfon. 



II 




f'l. AB=12 in.=e, th • 
edge of the cube. 

2. AN=AM=\AB = 

6 \w.-=\e. 

3. .-. MN=ML=MO 
=V(AN-+AM*) 
=\/(2AN 2 )= 
AN\/2=\y/2e, the 
side of the hexagon, FIG. 77. 

4. MQ=\ML=\ onV2e=^\/2e. Then 

5. 0R=V(M0*-MQ'>)=V[(W i ±ey~— (W^) 2 ]=W^ 
the radius of the circle. 

6. .••. 20P=2x(W^ e )=W^=W^X^=W^= 
14.6969382 in., the diameter. 

III. .*. The diameter of the largest circular ring that can be 
put in a cubical box whose edge is 1 foot, is 14.6969382 in. 



I. A fly takes the shortest route from a lower to the opposite 
upper corner of a room 18 feet long, 16 feet wide, and 8 feet 
high. Find the distance the fly travels and locate the point 
where the fly leaves the floor. 



MENSURATION. 



305 



Construction. — Let FABE — D be the room, of which AB is 
the length, AF the width, and 
AD the height; and let F be the 
position of the fly, and C the op- 
posite upper corner to which it 
is to travel. Conceive the side 
ABCD to revolve about AB 
until it comes to a level with the 
floor and takes the position of 
ABCD'. Then the shortest 
path of the fly is the diagonal 
FC of the rectangle FD'C'E, 
and P will be the point where 
the fly leaves the floor. 




A, 



II. 



1. AB=a=lS ft., the length of the room, 

2. AF=b=lQ ft., the width, and 

3. AD=/i=S ft., the height. 

4. FZy=FA+AZy=6+fi=lQ ft.+8 ft.=24 ft. 

5. FC / =\/(EI) /2 J r lJ / C / ' 1 )=V[(o+/i) 2J ra^] 



Then 



B. 



=V / [(16+S) 2 +18 2 ]=30 feet, the length of the 
path of the fly. 

FZy.D'C ::AF:AP, from the similar triangles 
a IT Fund PAF, or 

ab _ 18X16 
~'b+,~ 16+8 
=12 feet, the distance from A to where the fly 
leaves the floor. 



b+/i 



b:AP. Whence, AP-- 



III. 



30 feet is the distance the fly travels, and [floor. 

12 feet is the distance from A to where it leaves the 



Remark. — If we conceive the side BCHE to revolve about 
EH until it is level with the floor, the path of the fly will be 
FC" and the length of this is V[0+/;) 2 +£ 2 ]. But V[(*+ A ) 2 
+^ 2 ] ^> V[(^+^) 2 "f" <22 ]> because, by expanding the terms under 
the radicals, it will be seen that the terms are the same, except 
lah and 2b/i, and since a is greater than b, FC is less than FC". 
When a=b, FC'^FC". 

I. How many acres are there in a square tract of land con- 
taining as many acres as there are boards in the fence inclosing 
it, if the boards are 11 feet lonsf and the fence is 4 boards hisfh? 



. (side)" . . . . . , , . 

1. =n umber ot acres in the tract, the side being ex- 
100 

pressed in rods. 

2. 4xKHX-s"^=number of feet in the perimeter of the 

field. 



306 



FINKEL'S SOLUTION BOOK. 



4. 



• 4x L — n — J 

inclosing the tract. 



•number of boards in the fence 



(side) 



=4 



c 



4xl6£X«dfe 



- 1=24 X^e. Whence, 



160 L 11 

5. (szde) z =160X^4:Xside=SSA0Xstde. 

6. \ side=3840 rods=12 miles. 
^7. .'. (3840) 2 -r-160=92160=number of acres in the tract. 

III. .-. There are 92160 A. in the tract. 

(Milne s Pract. Arith., p. 362, prob. 7J. 

SECOND SOLUTION. 



1. 16=number of acres comprised between two panels of 

fence on opposite sides of the field. 
2 lA.=43560sq. ft. 

3. 16 A.=16X43560 sq. ft.=696960 sq. ft. 
\. llft.=the width of this strip comprised between the tvvo 
II. <^ panels. 

t. .-. 12 mi. =63360 ft.=696960-KLl, the length of the strip, 

which is the width of the field. 

6. 144 sq. mi.=( 12 ) 2 =the area of the field. 

7. 1 sq. mi.=640 A. 

8. 144 sq. mi.=144x640A=92160A. 

III. .-. There are 92160 A. in the tract. 

Explanation — Since for every board in the fence there is an 
acre of land in the tract for 4 boards, or one panel of fence there 
would be 4 A. Now a panel on the opposite side of the field 
would also indicate 4 A. Hence, between two panels on oppo- 
site sides of the field there would be comprised a tract 11 ft. wide 
and containing 8 A. But this would make boards on the other 
two sider of the field have no value. Now the boards on the 
other two sides having as much value as the boards on the first 
two sides, it follows that we must take twice the area of the 
rectangle included between two opposite panels for the area com- 
prised between two opposite panels in the entire tract. Hence, 
between two opposite panels in the tract there are comprised 16 A. 
The length of this rectangle is 16x43560-^1 1=63360 ft =12 mi., 
which the length of the side of the tract. 

In any problem of this kind, we may find the length of a side 
in miles, by multiplying the number of boards in the height of 
the fence by 33 and divide the product by the length of a board, 
expressed in feet. 



MENSURATION. 307 

I, How many acres in a circular tract of land, containing as 
many acres as there are boards in the fence inclosing it, the fence 
being 5 boards high, the boards 8 feet long, and bending to the 
arc of a circle? 

Construction. — Let C be the center of of the circular tract, AB 
—A C=B, the radius, and the arc AB=& feet. Then the area of 
the sector is 5 A. =2 17800 sq. ft. 

'1. 5A.=5x43560sq. ft.=217800 sq.ft., the area of the 

sector ABC. 
2. £(i4Lffx4C)=4(8X^C)=4^C=area of the sector 
ABC. 
II. v3. •*• 44 C=2 17800 sq. ft. Whence, 

4. 4C=217800-7-4=54450 ft.=3300 rods, the radius of the 
circle. 

5. ..*'. it X ( 3300 ) 2 -f- 160=68062. on = number of acres in 
tract. 

II. .-. There are 68062.5 n A., in the tract. 

I. What is the length of a thread wrapped spirally around a 
cylinder 40 feet high and 2 feet in diameter, the thread passing 
around 10 times? 

1. 2rr ft.=ABCA (Big. 79), the circumference of the cyl- 
inder 

2. 4 ft. =40 ft.-f-10=4/s the distance between the spires. 

3. \/[(27r) 2 +4 2 ]=2v/[> 2 +4] ft.=ABE, the length of 
one spire. 

4. .*. 10x2 V / i> 2 +4]ft. =20V|> 2 +4] ft.=74.4838 ft., the 
entire length of the thread. 



ii. 



■& 



III. .-. The entire length of the thread=74.4838 ft. 

Remark. — Each spire is equivalent to the hypotenuse of a right 
angled triangle whose base is ABC A and altitude AB. This 
may be clearly shown by covering a cylinder with paper and 
tracing the position of the thread upon it. Then cut the paper 
along the line AFK and spread it upon a plane surface. AEF 
will then be seen to be the hypotenuse of a right-angled triangle 
whose base is A CBA and altitude AB. 

I. A thread passes spirally around a cylinder 10 feet high 
and 1 foot in diameter. How far will a mouse travel in unwind- 
ing the thread if the distance between the coils is 1 foot? 

Construction. — Let A CB — Khe a portion of the cylinder and 
ADEB^GKa. portioa of the thread. Let A be the position of 
the mouse when the unwinding begins, B its position at any 
time afterwards, ABN a portion of the path it describes, and BD 
the portion of the thread unwound. Draw DC parallel <"0 HB 
and draw OD and OC. Then 



308 



FINKEL'S SOLUTION BOOK. 



IIJ 



1. AB=2R=1 foot, the diameter of the cylinder. 

2. a=10 ft., the altitude. Let 

3. 0=X.he angle AOC, 

4. s=AJV, the length of a por- 
tion of the curve, 

5. x=UM, and 

6. y=PP. Then 

7. PC=arc AC=R0, 

8. GM=R cos (9, 

9. ML=IP=CP cos Z CP/ 

=i?(9 COS (|7T— Z PC/) 

=7? 6 sin IP Cf=R0 sin 0. 

10. x=UM-\-ML=Rcos 6-\-Rd 
sin #, and 

11. y=PT=IM=CM—CI 
=RsinB— CP cos 6= 
Rsind—ROcos 0. 

12. dx=Rd cos Odd, by differ- 
entiating in 10, 

13. dy=R6s\nddd, by differ- 
entiating in 11. Now 

14. s=fy/[dx*+dy*\ 

15. .-. s=n(K0cosd l dd) 2 + 
(R6 sinO d6)*]*=R COdd 
^\R8\ But J FIG> 79: 

16. 6=27t, when one spire is unwound, and 

17. #=10x27r=207T, when the unwinding is complete. 
^18. .-. *=i^ 2 =!xK207r) 2 =1007r 2 =989.96044 ft, the 

distance the mouse travels to unwind the thread. 
III. .-. The mouse will travel 989.96044 ft. to unwind the thread. 

I. What is the length of a thread winding spirally round a 
cone, whose radius is R andaltitade a, the thread passing round 
n times and intersecting the slant height at equal distances apart? 

Let Pbe any point of the thread, (x,y, z,) the co-ordinates of 
the point ; and, let the angle PPC(=P>OC)=e, B0=a, the alti- 
tude, Z/0=R, the radius of the base of the cone, and r=the 
radius of the cone at the point P. Then the equations of the 




thread are : x 

27tn 
R 



r cos/9 (1), y=rz=r s:n# (2), and 2= 

(3). From the similar triangles DEP and BOB, 

r=— (a — z)=R\ 1 — - — i . . . (4). Now the distance between 
a v V 27tnJ v ' 

P and its consecutive position is \/ (dz 2 -\-dx 2 -\-dy 2 )= 

J['<£)'+<3)> - -AIL'+GD' 



MENSURATION. 



309 



J r f^l\ 2 "\dz (5). Substituting the value of r in (1) 

and (2),. and differentiating, we 
havedx= — - — cos0-f- 

(2nn—B) sin ')\dB and dy= — 

fsind— (2 nn— e)cosd~\dd. 



2rtn 



-df>. 



From (3), we have dz= 

v ' 2 Tin 

Substituting these values of dx, 
dy, and dz in (5), we have s= 

R°-\ cos6 +(2 an— fl)sinfl] 2 



a-" 



R 2 [sinfl — (2rtn— fl)cosft] 2 



<*0 




f/G. 50. 



R(2nn—Q) |r^ 2 -f/? 2 



2-i-/? 3 



+/? 2 (27rw— /?) 2 ]^ =j 



2 7T n 



( 2 7 r K -ff) 2 ]-7?(' , -!±^)log (! [(2^-W)+V / ((2^-^) 2 + 



^)]l*'"==W(«'+* , -H*VJe 

^ 2 • / - l 'o . <r 2 +/? 2 , rlnnRJ, 



) 



?i 1o g ,[ 



-7?_|_ V '(^2_|_7 ? 2_)_4 7r 2 ; ,27 ? ! 



±7t?zR 



] 



=iV(^ 2 +4* 2 « 2 i? 2 ) 



7; 2 



:loi 



. VK 2 +^ 2 ) 

p 27r^7?-fV(// 2 +47r 2 ^ 2 i? 2 ) -i 

where h=\/ (a 2 -\-R 2 ) , the slant height. 

Note. — This solution was prepared for the School Visitor, by the author. 

I. A thread makes n equidistant spiral turns around a cone 
whose slant height is /i, and radius of the base r. The cone 
stands on a horizontal plane and the string is unwound with the 
lower end in contact with the plane, the part unwound being 
always tense. Find the length of the trace of the end of the 
string on the plane. 

Let MHhe the part unwound at any time, //being the point 
in contact with the cone, and £M=zi, the trace on the plane up 
to this time. Put arc RjE=x, Af/=y, E being the point in the 
circumference of the base in the line AH. Let JVIhc the posi- 



310 



FINKEL'S SOLUTION BOOK. 



tion of the string at the next instant, D and /being homologous 
points with E and H. Draw HK parallel to ED. Then h : 
HK AK 



DE-,.AK,HK^^=^....(1) 



Now since the arc BE 



=x, is proportional to the distance the point 
of contact of the thread with the cone has 

ascended, x\h — y\ '.27trn'.h. or- = — =-=£. 

J 2rmt h 



dx 



2nrn 



— ... (2). This is negative 

dy y v ' 

since y decreases as x increases. It is evi- 

ED dx Inm. 

dent from the figure that --_ =■— -= = — 

s IK dy h 

By similar triangles, IK'.HKy. HE\ME, 

MP HFC 

that is, form (1) and (2), we get- = ^ 

ED y dx y ^nm 
^~IK X h = dy X k = ' ~^^ y 



Therefore, ME-. 



27trn . _ . 

--j2-( /i - y)y 



-y 

(3). 



(4) 



Put ME=t. Then ^ == _^!(/z_2v) 
dy h 2 

. . . . (5). By similar figures r\ME\ '.ED; 

- rn MExED „^ 2rtrn rTjr 

MP= - =—ME X — 7— X IK. 

r ft 

MP dv ±n 2 n-r 




FIG. 8h 



( ^) ,;:... (6 ), 



From (3), put MP=v, then _ 

7A ay 

Equation (5) gives the entire addition to the line ME which 

consists of JVP+ED, since PE=ME. Consequently, NP 

dt dx 2nrn .27rr?t Anm ff _. ^ T , r7 . TI> 

( h —2y)+—r-=-—y .... <7). Now MN 2 



dy dy 



1 r 1y1 



=MP 2 + JVP 2 in the limit. Therefore fpJ=J^£ 

^+-j^{h-yY) .... (3). V(8)=(9), ^=-^x 

^1 1-f- (^— JK) 2 ), the intregal of which is »., the length of 

h 2 Ar r h 

the trace. Put h — y=z, and— — ~—a 2 > Then u=-— I (h — z) 

•(,*+,*)*.,., (io). or«=^ + (g_g)v(^+^) 



+ 2rl 



° Se L 



^ + \/(« 2 +^ 2 ) 



] 



(11). Write for /i, its equal, 



MENSURATION. 



311 



A.f 2?* 
anrt, in (11) and we have (12), u=- • + — 



C 2 ^ 

I 7171 I 

V tin J 



\/(l + n 2 n 2 ) + 2r \og e [n?t -fV(l + 



)]• 



This result is independent of A, the cone's slant height, but 
involves ?z the number of turns of the thread. 

Note. — This solution is by Prof. Henry Gunder and is taken from the 
School Visitor, Vol.9, p. 199. Prof. Gunder stands in the very front rank 
of Ohio mathematicians. He has contributed some very fine solutions to 
difficult problems proposed in the School Visitor and the Mathematical 
Messenger. He is of a very retiring disposition and does not make any pre- 
tentions as a mathematician. But that he possesses Superior ability along 
that line, his solutions to difficult problems will attest. Prof. Gunder was 
born at Arcanum, O , Sept. 15th, 1837. He passed his boyhood on a farm 
and it was while following a plow or chopping winter wood, that difficult 
problems were solved and hitherto unknown fields of thought explored. He 
became Principal of the Greenville High School in 1867. After seven years' 
work here, he became Superintendent of the Public Schools of North Man- 
chester, Ind. After five years' work at this place he became Superintend- 
ent of schools of New Castle, Ind In 1890, Prof. Gunder was elected pro- 
fessor of Pedagogy in the Findlay, (Ohio) College. 

I. A woman printed 10 lbs. of butter in the shape of a right 
cone whose base is 8 inches and altitude 10 inches. Having com- 
pany for dinner, she cut offa piece parallel to the altitude and con- 
taining^ of the diameter. What was the weight of the part cut off? 

Constructio7i. — Let ARC — G be the cone, AC the diameter 
and OG the altitude. Let E be the point 
where the cutting plane intersected the the di- 
ameter, F the corresponding point in the slant 
height, and DLPKB the section formed by the 
intersection of the cone and the cutting plane. 
Through F pass a plane parallel to the base 
AR C and anywhere between this plane and 
the base, pass a plane NLMK. Then, 




A C=2R=8'm., the diameter of the base, 

OG=a=10 in., the altitude, and 

0E=0C—E C=R—$A C=R—IR=$R=:\$ in.=c,the 

distance of the cutting plane from the altitude. Let 
GQ=x, the distance of the plane MLNK from the 

vertex G. Bv similar triangles, 
OC\OG\\EC\EF, or R'.a'.'.R^c'.EF. Whence, 

■«*-*U| in. 



EF= 



R 



By similar triangles, 
GO\OC::GQ:QM, or a:R: :x'.QM. Whence, 
QM=LQ=—. Now, 



312 



FINKEL'S SOLUTION BOOK. 



9. area of LKM=area of LQKM—area of LKQ But 
10 



II, 



. area of LQKM=2 (jZQ 2 cos " 1 ^T)= (—} 



x 



-W- ) and 



11. «m ofZATG=i(L^xC/)=K2^/Xc)=£/Xc= 

12. .-. ^4r^« of the segment LKM= cos"" 1 ! — ) — 

(c-^a)i/(7?*x 2 — c 2 a 2 ). 

13. (^ 2 cos- 1 (£)-£ A /(7? 2 ^_. 2 , 2 ))^ = an el, 

merit of volume of the part cut off. 



14. .-. J 




)s ~ ] (£K^* 2 -^> 



-4a j ff'coB-^.g)— 2iV(^-^')+^X 



iog^ 



R+V(R' 



rj |=y |4«co^(i)--2X 



HV[4 2 -(ii) 2 ] + ,Vx42io ge [ 4+A/ [y ^ )2] ], 

±=y j 4»cotf-*(*.)-V\/2 + «log # [2+|V2] j , 

= 1 3° | 42 X M-t«l^- 6 9V2 + iflog,[2 + fv / 2] j , 
=V° ^ 19.6938154— 10.0562976 + .6396202 £ = 



III. 



34.223792 cu. in., the volume of the part cut off. 

15. ^7ri? ¥ =iXl0x4 2 X7r=53^7r cu. in., the volume of 

the whole cone. 

16. 10 lbs.=the weight of the whole cone. Hence, by pro- 

portion, 

17. 53^ cu. in. : 34.223792 cu. in. : : 10 lbs. : ( ?=2.04258 lbs. ) 

.-. The weigh of the part cut offis 2.04258 lbs. 



I. After making a circular excavation 10 feet deep and 6 feet 
in diameter, it was found necessary to move the center 3 feet to 
one side; the new excavation being made in the form of a right 
cone having its base 6 feet in diameter and its apex in the surface 
of the ground* Reqired the total amount of earth removed. 



MENSURATION. 



313 



Construction. — Let ABC — F be the cylindrical excavation 
first made, A C the diame- 
ter, HO the altitude. Let 
A be the center of the con- 
ical excavation, GAH its 
diameter, and AF\ an ele- 
ment of the cylinder, the 
altitude. Pass a plane at 
a distance x from O and 
parallel to the base of the 
excavation. Let figure II. 
represent the section thus 
formed, the letters in this 

section corresponding to FIG. 83 

the homologous points in the base represented by the same let- 
ters in the base of the excavation. An element of the earth re- 
moved in the conical excavation is (area BAKGNB)dx. The 
whole volume removed in the conical part of the excavation is 




£ 



(area BAKGNB)dx. For let 



1. HO 

2. HA 

conical parts 



:tf=10 ft., the altitude of the excavation, 

r=Z ft., the radius of the cylindrical and the 



rx 



3. AB—AN= — . This is found from the proportion of 

similar triangles. 

4. BI 2 = (rx-^a— AI)(rx^-a+AI). 

5. BI 2 =(2r—AI)AI. 

6. ... (2r—AI)AI=(rx-^a—AI)(rx-±-a+AI). Whence, 

7. AI= rx 2 S-2a*, 



Also 



8. BI=— -x/(4a*— * 2 ), 



2a 



9. HI-- 



rx- 



'2(7 



i)- 



Now 



10. area of BDAKGNB=2( area of BDAN+arca of 

NAG). But 

11. \n (r 2 x 2 -^-a 2 )= the area of the quadrant NAG, and 

12. area of BDAN=area of sector BAN-\-area of trian- 

gle HBA — area of sector BDAH. Now 

13. area of sector B A N=\ ABy s A B sin~^ (A I^-AB) 

= ( r 2 v 2^_ 2 a 2 )sin- 1 (-^-), 

14. area of triangle ABH=\(AHxBI)=\rX(rx^-- < la 2 ) 

XV{^ a *— x2 )=(r 2 x-±-±a 2 )\/(±a 2 —x 2 ), and 

15. area of sector B£>AH=ly[AHx AHcos- 1 (H/-+-BH)] 

=4r 2 X cos" 1 [1— (*-i-2a)]. 



314 



FINKEL'S SOLUTION BOOK 
•. Area of BDAKGNB 






\ 1 f x 2 * • 



2e : 



r 2 ,* 2 r 2 # 2 . /^ # "X r 2 # 



2^ 2 ' 
-r 2 cos-i 



(-£) 



"■ ■■■ r -X *!S?'+ £ S ! »»-'(s)+S« ,, --*'> 



■r^ cos" 



— I # z sin" 



(x 2 \ ) r 2 f a 

la* J ) 6 ~a 2 Jo 

(x A r 2 /*" /*« 

Sjlf+W. *V(4«'—*)*^/, cos- 

( i_ £)^ = * 7rar2 +5[* x3sin " i ^( x2 + 8a2) 

/^1 ^ 2 ^io/. 2 2^n s /64— 27V3— 2?r\ , 

=the volume of the cylindrical part of the ex- 
cavation. 
18. 7r«r 2 =the volume of the conical part. 

64— 27y'3— 2n\ „ ^64— 27V3+16*' 



>-(' 



) 



III. 



19. ,. ««r*+Q lg y _. y ig 

<zr 2 =337. 500554 cu. ft., the volume of the entire ex 
cavation. 

rp, . ,,, ,. /^64— 27V3+167TX 
.•. Ine volume of the excavation =1 ~ — ! 1 

or 337.50055+cu. ft., correct to the last decimal place. 



\ar 



Note. — This problem was proposed in the School Visitor by Wayland 
Dowling, Rome Center, Mich. A solution to the problem, by Henry Gun- 
der, was published in Vol. 9, No. 6, p. 121. The solution there given is by 
polar coordinates. The editor gives the answers obtained bv the contribu- 
tors; viz., Mr. Dowling, H.A.Wood, R. A. Leisy, and William Hoover. 
Their answers differ from Mr. Guilders and from each other. Mr. Gun- 
der's answer is 337.5-fcu. ft., the same as above. There is a similar problem 
in Todhunter's Integral Calculus, $. 190,prob. 29. A 

I. A tree 74 feet high, standing perpendicularly, on a hill- 
side, was broken by the wind but not severed, and the top fell di- 
rectly down the hill, striking the ground 18 feet from the root of 
the tree, the horizontal distance from the root to the broken 
part being 18 feet, find the height of the stub. 

Construction. — Let AD be the hill-side, AB the stump, BD 



MENSURATION. 315 

the broken part, and A C the horizontal line from the root of the 
tree to the broken part. Produce AB to ~E and draw DE paral- 
lel to A C. 

1. Let AB=x, the height of the stump. Then 

2. BD=74: ft. — x=s — x, the broken part, since AB-\-BD 
=74 feet. 

3. Let AD=a=M ft., the distance from the foot of the 
tree to where the top struck the ground, 

4. A C=3=18 ft. , the horizontal distance from t lie foot 
of the tree to the broken part. 

5. x=AB, the height of the stump. Then 

6. BC=\/(AB 2 +AC 2 )=^(x 2 +b 2 ) . . (1). Inthesim- 
lar triangles BA C and BED, 

7. \/(x 2 -\-b 2 ) : x : : s — x : BE. Whence, 

8" .**-$$) ■ ■ • - (2) ' A1S °' 
9. v '(*2_|_£2) , b , ..s—x.DE. Whence 



II. 



11. AE=BE—BA 



>j(x 2 +b 2 ) 

(4). 

12. AE 2 +ED 2 =AD 2 , or 

is. i.^) '■ i +i,! ( ^ ) j' «*«* 




=<z 2 . . . »(5). Developing (5), we have 

14. <[( s 3— a ~+b 2 )x*—4s(s 2 —a 2 +2b 2 )x*-{-(s++a*— 
2a~s-+$b 1 s 2 —Aa 2 b-)x 2 —Ab 2 s(s 2 —a 2 )x= 
—b*{s*—a*) ....(G). 

15. 1161* 4 — 91908* 8 +1959876* 2 ---25894080*+ 377913600 

=0 (7), by substituting the values of a, b, 

and s in (6). 

16. .*. x =24 feet, the height of the stump, by solving (7) 
bv Horner's method. 

III. .•. The height of the stump is 24 feet. 

Note. — This problem was taken from the Mathematical Magazine, Vol. 
I., No. 7, prob. 84. In Vol I., p. 184, of the Mathematical Magazine is a so 
lution to it, given by C. H. Scharar and Prof. J. F. W. Sheffer The solu- 
tion there given is different from the one above. 

I. What is the longest strip of carpet one yard wide that can 
be laid diagonally in a room 30 feet long and 20 feet wide ? 

Construction. — Let ABCD represent the room and EFGH 
the strip of carpet one yard wide placed diagonally in the room. 



316 



FINKEL'S SOLUTION BOOK. 



IlJ 



1. Let ^4^=^=30 ft, the length of the room, 

2. BC=b=20 ft., the width, and 

3. IZG=c=3 ft., the width of the carpet. Let 

4. BF=P/D=x. Then 

5. FC=AJ7=20— x=b— x. 

6. BF=\/(FF-—BF 2 )=\/(§—x z )=V(c' 1 ~x 2 )..n), 

7. AF=GC=AB— BB=a— \'(c 2 — x 2 ) .... (2). By 

similar triangles, 
EF-.BF: : GF: GC, or 



8. 
9. 

10. 



14. 



c : x : : GF 

Whence, 



a — ^(c 2 — x 




Again, we have 
BF\BB\:GF\FC, or 

c.hj(c 2 — x 2 )::GF: b — x, 

.(4). By equating G^in(3)and(4). 



FIG. 85. 



LrF= 



sj(c 2 —x 2 ) 
c(h—x) _ c [a—sJ(c 2 — x 2 )] 



sj(c 2 —x 2 ) 
15. bx— x 2 =asj(c 2 - 



...(5). 



-x 2 ) — c 2 -j-x 2 ...(6), by dividing (5) by 



cand clearing of fractions. 

16. c 2 — bx — 2x 2 =asJ(c 2 — x 2 ) . ..(7), by transposing in(6). 

17. 4* 4 — 4^ 8 + (fl 2 +3 2 — 4c 2 ) x*-\-26c*x=c*(a 2 —c*). % ;. 

. . . (8), by squaring (7) and transposing and com- 
bining. 

18. 4* 4 — 80* 3 +1264* 2 +360*=8019 (9), by restoring 

numbers in (8). 
.-. #=^2.5571+ft, by solving (9) by Horner's method. 



• >V /( C 2_*2 ) = y(9_^2 ) = i.5689 ft. Then, 



III. 



19 
20 

21. GC=30— \/(9— ^ 2 )=28.4311 ft, and 

22. FC 20 x=l 7.4429 ft. 

23^ .-. GF=\Z(FC 2 +GC 2 )=s/[(28A311) 2 -\-(17A229) 2 ] 
=33.3554 ft, the length of the carpet 

.-. The length of the strip of carpet is 33.3554 ft. 



I. What length of rope, fastened to a point in the circumfer- 
ence of a circular field whose area is one acre, will allow a horse 
to graze upon just one acre outside the field? 

Construction. — Let ABPC be the circular field and P the 
point in the circumference to which the horse is fastened. Let 
BP represent the length of the required rope. Draw the radius 
B O of the field and the line B C. Then 

' 1. 1 A.=160 sq. rd.=the area of the field ABPC, and 

2. BO=OP=R=\/{im-±-7t)=A J(^Y the radius of 



MENSURATION. 



317 



II 



9. 



10. 



11. 



12. 



13. 



14. 



20. 



Hence, 



length 



of the 



the circular field. Let 
0=the angle BPO=the angle OB P. 
tt— 20=the angle BOP. Now 
BP=AP cos Z APB=2Bcos&, the 

required rope. The 
area BPCD over which the horse grazes: 

BE CDB—areaBE CPB. 

But 
area of circle BE CD= 

7tBP 2 =7r 4P 2 cos 2 0= 
47tP 2 co& 2 6, and the 
« 7-rrt! HE CP= 2 X ( 0^# of 

sector E PB-\-area of seg- 
ment BPH). Now 
rtretf of sector EPB=\BPx 

arc BE=^X2Bcos f i X 

27?cosVx#=2/t ,2 flcos 2 9, and 
#/var of segment BPJP=area of sector BOP — tfr^a: of 

\\\wcr}<; OBP=±BOxa>'cBHP—±OPxBF= 

} £ [RXR(tt— 20)]— £/?X/?sin(?r— 20), 

=fff 2 (7r— 20)— |i? 2 sin20. 
.-. yW J g , ^ , Ci9=2[2i? 2 ^cos20+i/?^ T __2^)— 

fff 2 sin29]=i? 2 [40cos 2 0+7r--20— sin20]== 
l+cos20' 




7? 



[40(y^M?^ + ^_20-- s in20l==i?«[ 7 r+ 



:4tt7? 2 cos 2 6>— i? 2 [tt +2 <9cos2^— 



2#cos2#— sin 2 2#]. 

sin2#]. But 
7T/? 2 =1A.=160 sq. rd.=the area of BPC&B, by the 

conditions of the problem. 
.-. 47ri? 2 cos 2 #— i? 2 [7T+2flcos2^— sin20]=7rA 72 . 

Whence, 



15. 4*f 1+COs2 * > \— [^+20cos20— sin2fl]=rr ; 



v— 2~J 



or 



16. 27r+27rcos2r9 — 7t— 26 cos2# +sin2 6=n. 

17. .-. 2 0cos2^— sin20=2wcos20,or 

18. 26 — tan2#=27T, by dividing by cos2#. Whence, 

19. 0=51° 16' 24", by solving the last equation by 

method of Double Position. 



the 



'. A , / > =27?cos^=8jr^ ) ^cos"=8.92926+rods. 



III. .'. The length of the rope is 8.92926+rods. 

I. If a 2-inch auger hole be bored diagonally through a 4-inch 
cube, what will be the volume bored out, the axis of the auger 
hole coinciding with the diagonal of the cube? 

Formula. — I =r 2 ^/3( nc — 2^2), where eis the edge. 



318 



FINKEL'S SOLUTION BOOK. 



Construction. — Let AFGD be the cube and DF the diagonal, 
which is also the axis of the auger hole. The volume bored out 
will consist of two equal tetrahedrons acd — D and efg — F plus 
the cylinder acd—f, minus 6 cylindrical ungulas each equal to 
ace — b. Pass a plane any where between gaud b, perpendicular 
to the axis of the cylinder, and let x be the distance the plane is 
from £>. Now let 



II. 



1. AB=e—k inches, the edge of the cube ; 

2. DF=\/3s—4\/3, the diagonal of the cube; and 

3. r=± inch, the ra- 

dius of the auger, 
or the radius of 
the circle acd. 

4. ac—ad= dc = ry3 

5. Z>c = JrV6 = -|\/6 5 

by the similar tri- 
angles dDc and 
HBc. 

4(W6) 2 -^ 2 ]= 
^y'2==-|y2 5 the al- 
titude of the tetra- 
hedron acd — D. piQ gj 

7. .-. 2v-= %(area of base X altitude) = 2(^3 X «£ 2 XiX 
-i-r\/2)=;^6r 3 =^6, the volume of the two tetrahe- 
drons, 

8. v / =7tr 2 X(^ > F~2 times the altitude of acd—F>) = 

izr 2 (e\/Z — fcV%)~*(W&-rW2h the volume of the 
cylinder acd — f. 

9. be=^r\/2, by similar triangles, not shown in the figure. 
10. ^r\/2-f— |r\/2=r\/2=distance from D to where the 

auger begins to cut an entire circle. 

1 — ±x\/2=versin of an arc of the ungulas at a distance 
x from D. 



13. 




11. 

12. 2rcos 



( - l=an arc of the ungulas at a distance 

D. 

- J — \xsl2(r 2 — \x 2 ) =the area of a seg- 
ment at a distance x from D. 



x from D 

i-2 rr^c 1 



14. .\<fr"=6 



r rV2 [^2r2cos - 1 Q^^x s !2{r 2 —\x 2 ) h ~\dx 



MENSURATION. 



319 



WK 



|-,V2r 

± v 2\ 2 






r3(|v/6— 7rV2). 
15. .'. F,the volume bored out,=2v-\-v'—6v"=^j6r s -\- 

7rr 2 (^V3— |rV2)-r 3 (| v 6— /t v 2) =r-v / 3(7rc — 2rV2) 
=16.866105 cu. in. 

III. .*. The volume bored out is 16.866105 cu. in. 

I. A horse is tethered to the outside of a circular corral. The 
length of the tether is equal to the circumference of the corral. 
Required the radius of the corral supposing the horse to have the 
libert} of grazing an acre of grass. 

Construction. — Let AEFBK be the circular corral, AB the 
diameter, and A the 
point w here the 
horse is tethered. 
Suppose the horse 
winds the tether 
around the entire 
corral; he will then 
be at A. If he un- 
winds the tether, 
keeping it stretched, 
he will describe an 
involute, APGH\ 
to the corral. From. 
H' to H, he will de- 
scribe a semi-circle, 
radius AJF=AJ7= 
to the circumference 
of the corral. From 
H through G to A, 
he will again de- 
scribe an involute. 

Then the area over which he grazes is the semi-circle HLH'+ 
the two equal involute areas AFGHA and AKGH'A+W\t. area 
BFGKB. 

Let C be the center of the corral and also the origin of co-ordi- 
nates, A G the *-axis and P any point in the curve APGH'. 

( 1. Let0=the angle A CE that the radius CE perpendic- 
ular to PE, the radius of curvature of the curve 
APGH' , makes with the x-axis, 

2. o =the angie AFEBK that the radius CK makes 
with the .Y-axis when the radius of curvature PE has 
moved to the position KG\ 

3. P=A C, the radius of the corral; 

4. p=PE=arc AFE=RV, the radius of curvature of 




FIG. 88. 



320 FINKEL'S SOLUTION BOOK. 

the involute ; 

5. x=CM and 

6. yz=PM, the co-ordinates of the point P\ and 

7. x =C Gand 

8 i/ — Q the co-ordinates of the point G. Then we have 
9. x=CM=lE—CD=PE(=arcAFE) cosl/EP, 

= LPEC—i OEC{=lECD),— CE cos/ EGD 
=R0 cos( I lEP—i E CB)—R cos( tt—0)=R6 cos 

[ \rt—( 7T— ^ )] — /? cos( 7t—6) = R6 cos ~{\n— 6) 
-R cos( tt—0)=RO cos0+/? sin6> . . . . ( 1 ). 

10. y=PM=PI-\-IM{=DE)=PE sinZ PEI-{-ECx 
s\nAECD=R0 *\v(V—i7r)-{-R sm(7z—f>)= R sin# 
— /?tf sin^ .... (2). When ^=^ =angle AFEBK 

11. ^^CG^/Pcos^+iP^sin^.... (3), and 

12. yo== Q=j? s:nfi —Rd o cosV o .... (4). Hence, from (4), 

13. "6> =^ sin <V^cosfl :=tan# .. . .(5). Then, from (3), 

11 14. ^ =7?cos^ +i?tan^ sin^ =7?rco^„+^^sin^ > ) 

V. COS(7 S 

=^eT R sec "° =iiv( i+ t;i » 2 " o)=i?V( i+ *.' )• • • • 

(5). Now 

15. BFGKB=<l\\KGKKC— sector BCK]=R* 6 —fi 2 
(0.-*)....(7). 

16. AFGHA+AKGH'=ZjdA=<lj\p 2 dd=\R 2 6 2 dd 

= iR*(S7r— 61).... (8), and 

17. HH / L=\7i(AHy=\7i(<lnRy=Z7t*R 2 ....{§). Ad- 
ding (7), (8), and (9), 

18. R 2 d — R 2 (6— 7r)-fi/? 2 (87r— # 3 )+27r 3 i? 2 = 
J? 2 ^-!- 1 -^^ 3 — -i-0 o 3 )=area over which the horse 
grazes. 

19. 1 A. =160 sq. rd.=43560 sq. ft.=the area over which 
the horse grazes. 

20 . . R 2 (7r-\- l ^7t^— ^^ 3 )=43560 sq. ft. Whence, 

«^JGtoS£p0 •■■•("»■ - 

22. =4 494039=264° 37' 18".35 by solving (5) by the 
method of Double Position. 

23. .-. i?=19. 24738 ft, by substituting the value of o in(lO). 

III. .-. The radius of the corral is 19.24738 ft. 

A 20-foot pole stands plumb against a perpendicular wall. A 
cat starts to climb the pole, but for each foot it ascends the pole 
slides one foot from the wall; so that when the top of the pole is 
reached, the pole is on the ground at right angles to the wall. 
Required the equation to the curve the cat described and the 
distance through which it traveled. 



MENSURATION. 



321 



Construction. =Let A C be the wall, P the position of the cat 
at any time, and B C the position of the ladder at the same time. 
Draw AP and to the middle point D of AP draw BD. Then 
AB=PB. 

<l. Let BC=20 ft=a, the length of the. ladder, 

2. AP=r, the radius vector of the - 
curve the cat describes, and 

3. 0=the angle PAB. 

4. 7t — 2#=the angle ABP, because 
the angle PAB— the angle 
BPA. 

<A -\h. AB=B C cos AB C=a cos(tt— 20) 
= — a cos2 6 , 

6. %AP=hr=AD=AB cos^ BAD 
= — a cos2#cos;9. 

7. .". r= — 2<?cos2frcos#, or 

8. r-|-2<2COs2# cos#=0, the equation of the curve de- 
scribed by the cat. 

1. Let 5=the distance through which the cat traveled. 

s=JV(<tr 2 +r 3 d0 2 )=2a / N (l— 12 cos* $ + 
44 cos 4 0—32 cos 6 d)dd, 
,u, = — a J *J(2 — 4 cos0 — cos 2 -f- 4 cos 3 cp)d<p. 




n. 



where 0=7r — 20, 
4. = — \a I v(6— 4 cos0— 2 cos 2 0-f 4 cos 3 0)^/0 



III. 



=1.1193 «, 

5. =22.386 ft., the distance through which the cat 
travels. 

\ r-|-2tf cos2 cos 0=0, is the equation or the curve, and 
\ 22.386 ft.=the distance through which cat traveled. 

Note. — The integration in this problem is performed by Cote s' Method 
of Approximation. 

I. Suppose W. A. Snyder builds a coke oven on a circular 
bottom 10 feet in diameter. While building it, he keeps one end 
of a pole 10 feet long, always against the place he is working 
and the other end in that point of the circumference of the bot- 
tom opposite him. Required the capacity of the oven. 

Construction. — Let AB be the diameter of the base and C G 
the altitude. At a distance x from the base pass a plane inter- 
secting the oven in F and E. Draw AE and A C. 



322 



II. 



FlNKEL'S SOLUTION BOOK. 

AB=2R=10 feet, the diameter of the base. 
A C=AE=2R=10 feet, by conditions of the problem. 
CG=y/{AC 2 — AG 2 )=(4R 2 — R 2 )=R>JS, the alti- 
tude. 

BH^=x' 2 =(SA G-\ GH) ( GB— GH)=ZA G 2 — 
2+Gx GH=%R 2 — 27?X CH(=EI), because EHis 
the ordinate of a semi-circle whose diameter is 2AB. 
From this, we find 



EI=^{4R 2 — x*)— R. 

7rEI*=7i[*/(4R 2 —x 2 )- 

area of the circle whose center is/. 



Then 
-R] 2 , the 



■Rydx= ( 

Jo 



?r[^/(4R 2 - 
-x 2 )]dx, 




—2R V (4R< 

8. =\3R 2 x~ i^ 2 - 4i? 3 sin-! -1 

— Rx\/(4R 2 — x*)\ , 

9. -=^7r7?2( 9V 3_ 47r)=4ir5 8 (9V3— 4*), 

10. =t»125(9V3—47r) =395.590202+ cu. ft. 
III. .-. The capacity is 395.490202 cu. ft. 

I. At each corner of a square field whose sides are 10 rods, a 
horse is tied v/ith a rope 10 rods long; what is the area of the 
part common to the four horses? 

Construction. — Let ABCD be the field and EFGH the area 
common to the four horses. Join EF, 
EG, GH, and EH. Draw JDK per 
peudicular to EF and draw DE and 
DF. Since AF=DE=I)K= GF= 
CE, the triangles ADF and EDC 
are equilateral and, consequently, the 
angle ADE=L_ADC— LED C = 90° 
— 60°=30°. Also the angle FCD= 
30 c . Hence, EZ)F=S0 G . Now let 

FIG. 91. 
r l. AE>=EE>=a=10 rods. Then 

2. Area of sector EDF=\ EF>X arc EKF=±[aX{2na) 

3. Area of t riangle EDF=\EFy ^DK. But, 

4. ^^=4Va[2a— V(4a 2 — AF 2 )], by formula of Prob. 
XXII., =W(2—V3), and 

!U5. DK=s/YDE 2 — (±EF) 2 ]=ia>J (2+s/3). Hence, 

6. area of triangle EDF=±asl(2-\-sl2>) X W(2— V3)=i« 2 . 




MENSURATION. 



323 



7» ,\ Area of segment EE= T \7ra 2 — ia 2 = T ^a 2 (7r— 3). The 

8. area of square EEGJ/=EE 2 =a 2 (2— V3). Hence, 

9. area of the figure EEG/f=a 2 (2— \/3 ) +4 X tV « 2 ( 7r ~ 3 ) 
=a 2 (^7t-\-l — ^3)=31.5147 sq. rd.=the area common 
to the four horses. 

III. .'. The area of the part common to the four horses is 
31.5147 sq. rd. 

Note. — This problem is similar to problem 348, School Visitor , to which 
a fine trigonometrical solution is given by Prof. E. B. Seitz. 

I. What is the length of the longest straight, inflexible stick 
of wood that can be thrust up a chimney, the arch being 4 feet 
high and 2 feet from the arch to the back of the chimney — the 
back of the chimney being perpendicular? 

Construction. — Let PDEC be a verticle section of the chim- 
ney, PB the height of the arch, PE the distance from the arch 
to the back of chimney, and APD the longest stick of wood that 
can be thrust up the chimney. 



II. 



1. Let PB=a=4: feet, the height 

2. PE=6=2 feet, the width of the 

chimney, 

3. #=the length of the longest stick 

of wood, and 

4. fr=the angle DA C. Then 

5. AP=PB cosec6=a cosecfl, 

6. PD=PE sec6=b sec 6. 

7. .-. x=AP+PZ)=a cosectf +b 

seed (1). Differentia- 
ting (1), 

8. 0= — a cos #-J-sin 2 0-\-b sin#-r- 



of the arch 



cos 2 <? 
;). a cos 3 fl= 



.,.(2), or 

--b sin 3 6 .. 



(3), by 




FIG. 92. 



clearing of fractions and transposing in (2). 



10. 



11. 



cos 3 # 
tanfl= 3 



==tan ; 



= T . Whence, 
o 



12. cottf 



From (3), we may also have 



-. Now, from trigonometry, 



13. \/(l+tan 2 )=sec0, and 

14. \ / (l-\-cot 2 0)=cosec8. Hence, by substituting in (1), 

15. ^=^ v / (l+cot 2 ^)+V(l+tan 2 0)= 

+glV(af+rf)==(«f+ < 5l)(V a i+^i)==(«S+3i)i J 

=V / [(« l +^) 3 ]=A/[(^+2t) 3 ]=8.323876+ft. 



324 



FINKEL'S SOLUTION BOOK. 



III. .-. The length of the longest stick is 8.323876-fft. 

J. A small garden, situated in a level plane is surrounded by 
a wall having twelve equal sides, in the center of which are 
twelve gates. Through these and from the center / of the garden 
12 paths lead off through the plane in a straight direction. From 
a point in the path leading north and at a distance of 4 furlongs 
47-jVi - yards fromthecenter of the garden, A. and B. start to travel 
in opposite directions and at the same rate. A. continues in the 
direction he first takes; B. 5 after arriving at the first road (lying 
east of him) by a straight line and at right angles with it, turns so 
as to arrive at the next path by a straight line and at right angles 
with it and so on in like manner until he arrives at the same 
road from which he started, having made a complete revolution 
around the center of the garden. At the moment that B. has 
performed the revolution, how far will A. and B. be apart? 



C,£>,E,F, G,H,I, 



Let O be the center of the garden, A the point in the path 
leading north from which A. and B. start, 
K,L, M, N, P, the points at which 
B. strikes the paths. The triangles 
OCA, ODC, OED, OEE, &c, are 
right triangles, OCA, ODC, OED, 
OEE, &c, being the right angles. Let 
S in the prolongation of AC denote the 
position of A., when B., arrives at P. 
It is required to find the distance AS. 
Let OA=a=4 furlongs, 47 ¥ 1 A 3' d -> AS 
= A C+ CDA-DE+ . . . +NP=x, PS 
=y, AP=z, n=12, the number of paths 

and lAOE= lCOD= /_DOC= 

Z JVOP=360°-+-7z=30 G . Then from the 

right triangles we have OC=OAx FIG. 93. 

cosAOC=acos0, OD= O C cos COB=a cos 2 0, OE=OZ?X 
cosZ?OE=acos 5 d, OP=OJVcos NOP=a cos*0; AC=OAX 
sin A O C=a sin0 , CD= O C sin C OD=a sin cosV, DE= 
jDOsmE>OE=asindcos 2 e,J\rP=ArOsmJVOP=^asiudcos n - i e. 
.'. z=OA — OP—a(\ — cos*0),and x=a sintf+a sin0cos0-f- 

a sin0 cos 2 0-f -\-a sin0 cos w-1 0=asin0(l-|-cos0-|-cos 2 + 

cos 3 0-f +cos™- 1 ^)=tfsin#(l-f-cos^)-r-(l— cos^)= 

a cot*0(l— cos»0). Hence, since lPAS=(90°+6 ), we have 
y=T/\x 2 +z*—2xzXcos(90 o +d )]=« cosec £0(1— cos"0) X 

V(l+sin 2 0)=^W-°X^^[l-(f) 6 ]xf ==3292 yd., nearly. 




Note -—This problem was proposed in the School Visitor, by Dr. N R. 
Oliver, Brampton, Ontario. The above elegant solution was given by Prof. 
E. B. Seitz, and was published in the School Visitor, Vol. 3, f. 36. 



MENSURATION. 



325 



I. A fox is 80 rods north of a hound and runs directly east 
350 rods before being overtaken. How far will the hound run 
before catching the fox if he runs towards the fox all the time, 
and upon a level plain? 

Construction. — Let C and A be the position of the hound and 
fox at the start, P and m corresponding positions of the hound 
and fox any time during the chase, and P' and n their positions 
the next instant, B the point where the hound catches the fox 
and CPP / B the curve described by the hound. Join m and P, 
and n and 7^; they are tangents to the curve at P and P / . Draw 
Pd and P'e perpendicular to AB, mo perpendicular to P' 'n , 
and P'p perpendicular to Pd. 



II. 



Let AC-- 

AB=6= 

Am=x, 

Bd=y, 

Pm=iv. 



=a=80 rds. 
:350 rds., 



6. arc CP=s, 



curve CPB=s ly and 




5. r=ratio of the FIG 94. 

hound's rate to the fox's. Then we have 
9 4 m?z=dx, 

10. ed=P / p=dy, 

11. PP'=ds, 

12. no—PP'=dw 

13. s=rx ... .(2). 

14. ds—rdx. Whence, 

dx 1 

15. y — . From the similar right triangles PpP / and ?non, 



.. (1), and 
From (2), we have, by differentiation 



ds r 

we have 
PP / [ mn;;pP'\ mo, or ds : dx 
dxXdy dy . dx 



dy 



ds ? 

■ ds=dw, or 



=— , since 



ds 



1G. 
17. 

18. 

r 

19. dy—r 2 dx=rdw .... (3). 

20. y— r 2 x=rw+C. ... .(4). 

and W=«, 

21. 0=ra-\-C\ Whence, C= — ra. 

22. .\_y — r\v=nt'-(-C=rw — r# .... (5). 

=3, and w=0, and (5) becomes 
3 — r 2 b= — «r, or r 2 b — ra=b. Whence 



iy : Dio. Whence, 
-. Substituting in (1), 



Integrating (3), 

But, since when x=0,y 



When x=b, v 



=0, 



23. 
24. 

25. 



'— r +U' 



" h 4^ 2 U- 



326 



FINKEL'S SOLUTION BOOK. 



26 - Hh^ g '+ 4 * 1 ' 

27. 2n3—fl=:V / a 2 4-4£ 2 , 

28. r3=i(«+Va 2 +4^ 2 ). But 

29. rb=s 1 , what (1) becomes when b—x. 

130. .-. s 1 =i(a-\-^Ja 2 +43 2 =392.2783 rods, the distance the 
hound runs to catch the fox. 



Note. — This solution is substantially the same as the one given by the Late 
Professor E. B. Seitz, and published in the School Visitor, Vol. IV, p. 207. 
The path of the hound is known as the "Curve of Pursuit." 

I. A ship starts on the equator and travels due north-east at 
all times ; how far has it traveled when its longitude, for the 
first time, is the same as that of the point of departure? 

Let B be the point of the ship's departure, PPJVits course, P 
its position at any time and iVits position at the next instant. 
Then PN is an element of the curve of the ship, which is known 
as the Loxodrome, or Rhumb line. Let 6=PP=the longitude 
of the point P, (p=PP=the corresponding latitude, (x,y, z) 
the rectangular co-ordinates of P, and gj—^7T=the constant 
angle PNO. 

Then we have for the 
equations of the curve, 
x=P G cosd 
=rcos0cos# .... (1), 
y=P G sin (J=rcos(pX 
sin# .... (2), and 
z=r sin0. .. .(3), where 
r is the radius of the 
earth. Now an element 
of a curve of double 
curvature, referred to 
rectangular co-ordinates 
is ^(dz*+dy*^-dx*). 
.-. PN=ds 
=<J(dz 2 +dy*+dx' 
...(4). 

(1), (2), and (3), 
dx= — r(cos#sin0<^0-|- 
coscpsinOdd), FIG- 95. 

dy= — r(sin#sin0^0 — coscpcosdd 6), and dz=rcos cpdcf). Sub- 
stituting these values in (4), </s=r\/[cos 2 0^0 2 -[-(sin Osirupdcp 
— cos0cos ddd ) 2 + (cos#sin0^0-f-cos0sin^6') 2 ]==rv / (cos 2 0^0 2 
+sin 2 0^0 2 +cos 2 0^ 2 )=rv / (^0 2 +cos 2 0^ 2 ) .... (4). Now 



Differentiating 




pQ=GPX-PQ=r cos<pd0 and JVQ=rd(p. 



MENSURATION. 327 

PQ coscpdd , ya J 

=tan<p. .-. ^j-k= — -TT— =tan9>, or cos0tfe/=tan 9<f0 .... (6). 

Substituting the value ofcos(pd8 in(5) ,ds=r\/ (dcfi 2 -{-tan 2 <pd(p 2 ) 

^(* -*»> ■ ■ ■ -< 7 >- B y inte ^ tin g w^V^S 

==tan<plog e [tan(i7r+-£0)] or ^ cot< P=tan( ^+10) . . . (8). 
Whence, 0=2tan- 1 (^ cot <P)—- J-tt . When #=2 7T and <p=^7T, 

0=2tan- 1 (^ 27r )— iar=89° 47' 9 // .6=.4988i-?r. .". te-^- X 

V 7 " y COS^7T 

(.4988i7r— ^)=rV2(.4988i7r)=2.21615937/'=8775.991093— mi., 
the distance the ship travels. 

The rectangular equations of the Loxodrome are V(^ 2 +J 2 ) 

\ -i y -iV } 

I e ai&n x-\-e atan x \ —2r, and x 2 -\-y 2 -{-z 2 =r 2 , where a=cot(p. 

The last equations are easily obtained from the figure. The 

first is obtained as follows: From (1) and (2), we find #=tan _1 - ; 

x 

also, x 2 -\-y 2 =r 2 cos 2 or co^<p=—sJ(x 2 -\-y 2 ). From (8), we get 

e fhotg> = _E^L = COS0 Whence, ^-^ + e^'^X 

l-|-sin0 1+V(1 — cos 2 0) ' 

\/(l — cos 2 0)=cos0. Transposing e9 coi( P, squaring, and reducing, 

we have cos(p(e 6co{( P-{-e~V coi(p )=2. Substituting the value cos0, 

{ -\ V x-l ) 

and 0, we have y/( x 2 -\-y 2 ) \ c aUin a>-|-£r« tan a > =2r. 

Note. — This solution was prepared by the author for problem 1501, 
School Visitor, but it was not published because of its difficult composi- 
tion. 



328 FINKEL'S SOLUTION BOOK 

PROBLEMS, 



1. What is the area of a field in the form of a parallelogram, whose 
length is 160 rods and width 75 rods? Ans. 75 A. 

2. Find the area of a triangle whose base is 72 rods and altitude 16 
rods. Ans. 1 A. 2 R. 16 P. 

3. Two trees whose heights are 40 and 80 feet respectively, s:and on op- 
posite sides of a stream 30 ft. wide. How far does a squirrel leap in jumping 
from the top of the higher to the top of the lower? Ans. 50 feet. 

4. How many steps of 3 feet each does a man take in crossing diagonal 
ly, a square field that contains 20 acres? Ans. 440 steps. 

5. Find the cost of paving a court 150 feet square; a walk 10 feet around 
the whole being paved with flagstones at 54 cents a square yard and the 
rest at 31% cents a square yard? Ans. $939.40. 

6. What is the area of a triangle, the three sides of which are respect- 
ively 180 feet, 150 feet, and 80 feet? Ans 5935-85 sq. ft. 

7. What is the area of a trapezium, the diagonal of'which is 110 feet, and 
the perpendiculars to the diagonals are 40 feet and 60 feet respectively? 

Ans. 5500 sq. ft. 

8. At 30 cents a bushel, find the cost of a box of oats, the box being 8 
feet long, 4 feet wide and 4 feet deep. Ans. $30.85%. 

9. Two trees stand on opposite sides of a stream 40 feet wide. The 
height of one tree is to the width of the stream as 8 is to 4, and the width of 
the stream is to the height of the other as 4 is to 5. What is the distance 
between their tops ? Ans. 50 feet. 

10. How many miles of furrow 15 in., wide, is turned in plowing a rect- 
angular field whose width is 30 rods and length 10 rods less than its diagoal ? 

Ans. 49% mi. 

11. The sides of a certain trapezium measures 10, 12, 14, and 16 rods 
respectively, and the diagonal, which forms a triangle with the first two 
sides, is 18 rods; what is the area? Ans. 163.796 sq. rds. 

12. Three circles, each 40 rods in diameter, touch each other externally; 
what is the area of the space inclosed between the circles ? 

Ans. 64.5 sq. rds. 

13. How many square *ncnes in one face of a cube which contains 2571353 
cubic inches? Ans. 18769 sq. in. 

14. Four ladies bought a ball of thread 3 inches in diameter; what por- 
tion of the diameter must each wind off to heve equal shares of the thread? 

First, .2743191 in. 
, \ Second, .3445792 in. 
jins. -j Thirdj .4912292 in. 

{ Fourth, 1.8898815 in. 

15. A gentleman proposed to plant a vineyard of 10 A. If he places the 
vines 6 feet apart; how many more can he plant by setting them in the 
quincunx order than in the square order, allowing the plat to lie in the form 
of a square, and no vine to be set nearer its edge than 1 foot in either case? 

Ans. 1870. 

16. Find the volume generated by the revolution of a circle about a 
tangent. • Ans. 2 / k 2 R z . 

17. How many feet in a board 14 feet long and 16 inches wide atone 
end and 10 inches at the other, and 3 inches thick? Ans. 45^ feet. 

18. If I saw through^ of the diameter of a round log, what portion of 
the cut is made? Ans. .196. 



MENSURATION. 329 

19. What is the surface of the largest cube that can be cut from a sphere 
which contains 14137.2 cu. ft.? Ans. 1800 sq. ft. 

20. Two boys are flying a kite. The string is 720 feet long. One boy 
who stood directly under the kite, was 56 feet from the other boy who held 
the string; how high was the kite? Ans. 717.8+feet. 

21. How many pounds of wheat in a cylindrical sack whose diameter is 
\)4, feet, and whose length is 1% yards? (ttz=3.1416) Ans. 417 .61 feet. 

22. How large a square can be cut from a circle 50 inches in diameter? 

Ans. 35.3553391 in. 

23. How many bbl. in a tank in the form of the frustum of a pyramid, 
5 feet deep, 10 feet square at the bottom and 9 feet square at the top? 

Ans. 107.26 bbl. 

24. From a circular farm of 270 acres, a father gives to his sons equal 
circular farms, touching each other and the boundary of the farm. He 
takes for himself a circular portion in the center, equal in area to a son's 
part, and reserves the vacant tracts around his part for pasture lands and 
gives each son one of the equal spaces left along the boundary. Required 
the number of sons and the amount of pasture land each has. 

Ans. 6 sons; 8.46079 A. 

25. At each angle of a triangle being on a level plain and having sides 
respectively 40, 50, and 60 feet, stands a tower whose height equals the sum 
of the two sides including the angle. Required the length of a ladder to 
reach the top of each tower without moving at the base. 

Ans. 116.680316+ft. 

26. If the door of a room is 4 feet wide, and is opened to the angle of 90 
degrees, through what distance has the outer edge of the door passed? 

Ans. 6.2832 feet. 

27. A tinner makes two similar rectangular oil cans whose inside dimen- 
sions are as 3, 7, and 11. The first hold 8 gallons and the second being 
larger requires 4 times as much tin as the other. What are the dimensions 
of the smaller and the contents of the larger? 

. \ Dimensions of smaller 6, 14, and 22 inches. 
' ( Capacity of larger 64 gallons. 

28. An 8-inch globe is covered with gilt at 8 cents per square inch; find 
the cost. A ns. $16.08. 

29. A hollow cylinder 6 feet long, whose inner diameter is 1 inch and 
outer diameter two inches, is transformed into a hollow sphere whose outer 
diameter is twice its inner diameter; find outer diameter. Ans. 3.59 in. 

30. A circular field is 360 rods in circumference; what is the diagonal of 
a square field containing the same area? Ans. 20.3 rods. 

31. What is the volume of a cylinder, whose length is 9 feet and the cir- 
cumference of whose base is 6 feet? Ans. 25.78 cu. ft. 

32. How many acres in a square field, the diagonal being 80 rods? 

Ans. 20 acres. 

33. How many cubical blocks, each edge of which is 3^3 of a foot, will 
fill a box 8 feet long, 4 feet wide, and 2 feet thick. Ans. 172S blocks. 

34. From one corner of a rectangular pyramid 6 by 8 feet, it is 19 feet 
to the apex; find the dimentions of a rectangular solid whose dimensions are 
as 2, 3, and 4, that may be equivalent in volume. Ans. 4, 9, and 8 feet. 

35.* A solid metal ball, 4 inches radius, weighs 8 lbs.; what is the thick- 
ness of spherical shell of the same metal weighing iy 8 lb., the external di- 
ameter of which is 10 inches? Ans. 1 inch. 

36. What is the difference between 25 feet square and 25 square feet? 

Ans. 600 sq.ft. 



330 FINKEL'S SOLUTION BOOK. 

37.* Find the greatest number of trees that can be planted on a lot 11 
rods square, no two trees being nearer each other than one rod? 

Ans. 152 trees. 
38.* A straight line 200 feet long, drawn from one point in the outer 
edge of a circular race track to another point in the same, just touches the 
inner edge of the track. Find the area of the track and its width. 

Ans. Area, 7ra 2 — 40007r sq. ft.; width, indeterminate 

39. The perimeter of a certain field in the form of an equilateral triangle 
is 360 rods; what is the area of the field? Ans. 543.552 sq. rd. 

40. A room is 18 feet long, 16 feet wide, and 10 feet high. What length 
of rope will reach from one upper corner to the opposite upper corner and 
touch the floor? Ans. 353 ft. 

41. How many bushels of wheat in a box whose length is twice its width, 
and whose width is 4 times its height; diagonal being 9 feet? 

Ans. 25 bu., nearly. 

42 Find the area of a circular ring whose breadth is 2 inches and inside 
diameter 9 inches. Ans. 69.1152 sq. in. 

43 * A round stick of timber 12 feet long, 8 inches in diameter at one 
end and 16 inches at the other, is rolled along till the larger end describes a 
complete circle. Required the circumference of the circle. 

Ans. 150.83 feet. 
44. A fly traveled by the shortest possible route from the lower corner 
to the opposite upper corner of a room 18 feet long, 12 feet wide and 10 feet 
high. Find the distance it traveled. Ans. 28.42534 feet. 

45.* From the middle of one side and through the axis perpendicularly 
of a right triangular prism, sides 12 inches, I cut a hole 4 inches square. 
Find the volume removed. Ans. 138.564064 cu. in. 

46.* Two isosceles triangles have equal areas and perimeters. The base 
of one is 24 feet, and one of the equal sides of the other is 29 feet. The 
area of both is 10 times the area of a triangle whose sides are 13, 14, and 15 
feet. Find the perimeters and altitudes. 

Ans. Perimeters, 98 feet; altitudes 35 and 21 feet. 

47. A grocer at one straight cut took off a segment of a cheese which 
had 3^ of the circumference, and weighed 3 pounds; what did the whole 
weigh? Ans. 33.023 lb. 

48.* A twelve inch ball is in a corner where walls and floor are at right 
angles; what must be the diameter of another ball which can touch that 
ball while both touch the same floor and the same walls? 

Ans. 3.2154 in. or 44.7846 in. 

49. What will it cost to paint a church steeple, the base of which is an 
octagon, 6 feet on each side, and whose slant height is 80 feet, at 30 cents 
per square yard? Ans. $64. 

50. A tree 48 feet high breaks off; the top strikes the level ground 24 feet 
from the bottom of the tree; find the height of the stump. Ans. 18 feet. 

51. How many acres in a square field whose diagonal is b% r °ds longer 
than one of its sides? Ans. 160.6446 sq. rd. 

52.* Three poles of equal length are erected on a plane so that their tops 
meet, while their bases are 90 feet apart, and distance from the point where 
the poles meet to the center of the triangle below is 65 feet. What is the 
length of the poles? A ns. 83.23 feet. 

53. A field contains 200 acres and is 5 times as iong as wide. What will 
it cost to fence it, at a dollar per rod ? Ans. $960. 

54.* What is the greatest number of plants that can be set on a circular 
piece of ground 100 feet in diameter, no two plants to be nearer each other 
than 2 feet and none nearer the circumference than 1 foot? Ans. 2173. 



MENSURATION. 331 

55. The axes of an ellipse are 100 inches and 60 inches; what is the dif- 
ference in area between the ellipse and a circle having a diameter equal to 
the conjugate axis? Arts. 600 tt=1884.96 sq. in. 

56. Find the diameter of a circle of which the altitude of its greatest in- 
scribed triangle is 25 feet. Arts. 33% feet. 

57. If we cut from a cubical block enough to make each dimension 1 
inch shorter, it will lose 1657 cubic inches, what are the dimensions? 

58. Show that the area of a rhombus is one-half the rectangle formed by 
its diagonals. Noble Co. Ex. Test. 

59. The length and breadth of a rectangular field are in the ratio of 4 
to 3. How many acres in the field, if the diagonal is 100 rods? 

60. A spherical vessel 30 inches in diameter contains in depth, 1 foot of 
water; how many gallons will it take to fill it? Holnies Co. Ex. Test. 

Ans. 39 gallons. 

61. A field is 40 rods by 80 rods. How long a line from the middle of 
one end will cut off 1)4> acres? Ans. 80.6 rd., nearly. 

62. A ladder 20 feet long leans against a perpendicular wall at an angle 
of 30°. How far is its middle point from the bottom of the wall? 

Ans. 10 feet. 

63. Four towers, A 125 feet high, B 75 feet, C 160 feet, and D 65 feet, 
stand on the same plane. B due south and 40 rods from A; C east of B 
and D south of C. The distance from A to C plus the distance from C to B 
is half a mile, and the distance from D to B is 82^£ yd. farther than the dis- 
tance from C to D. What length of line is required to connect the tops of 
A and D ? Ans. 240+rds. 

64. Find the volume of the largest square pyramid that can be cut from 
a cone 9 feet in diameter and 20 feet high? Ans. 270 cu. ft. 

65. A rectangular lawn 60yd. long and 40 yd. wide has a walk 6 ft. wide 
around it and paths of the same width through it, joining the points of the 
opposite sides. Find in square yards the area of one of the four plats in- 
closed by paths. Ans. 459 sq. yd. 

66. Which has the greater surface, a cube whose volume is 13.824 cu. ft., 
or a rectangular solid of equal volume whose length is twice its width, and 
its width twice its height? Ans. Cube, 424 sq. ft., more. 

67. The volume of a rectangular tin can is 3 cu. ft. 1053 cu. in.; its di- 
mensions are in the proportion of 11, 7, and 3. Find the area of tin in the 
can. Ans. 16% sq. ft. 

68. A conical well has a bottom diameter of 28 ft. 3 in., top diameter 
56 ft. 6 in., and depth 23 ft. 1.2 in. Find its capacity in barrels. 

Ans. 8023 bbl. 

69. A cylindrical vessel 1 foot deep and 8 inches in diameter was *§ full 
of water; after a ball was dropped into the vessel it was full. Find the di- 
ameter of the ball. Ans. 6 inches. 

70. Two logs whose diameters are 6 feet lie side by side. What is the di- 
ameter of a third log placed in the crevice on top of the two, if the pile is 9 
feet high?. Ans. 4 ft. 

71. Circles 6 and 10 feet in diameter touch each other; if perpendiculars 
from the center are let fall to the line tangent to both circles, how far apart 
will they be? A ns. 7.756 ft. 

72. What are the linear dimensions of a rectangular box whose capacity 
is 65910 cubic feet; the length, breadth, and depth being to each other as 
5, 3, and 2? A ns. 65, 39, and 26 ft. 

73. The perimeter of a piece of land in the form of an equilateral trian- 
gle is 624 rods; what is the area? Ans. 117 A. 13 31 P. 



332 FINKEL'S SOLUTION BOOK. 

74. Four logs 4 feet in diameter lay side by side and touch each other; 
on these and in the crevices lay three logs 3 feet in diameter; on these three 
and in the crevices lay two logs 2 feet in diameter; what is the diame- 
ter of a log that will lay on the top of the pile touching each of the logs 2 
feet in diameter and the middle one of the logs 3 feet in diameter? 

Ans. 

75. What will it cost to gild a segment of a sphere whose diameter is 6 
inches; the altitude of the segment being 2 inches, at 5 9 per square inch? 

Ans . 

76. A grocer cut off the segment of a cheese, and found it took | of the 
circumference. What is the weight of the whole cheese, if the segment 
weighed \% lbs ? Ans. 52.0228+lbs. 

77. Two ladders are standing in the street 20 feet apart. They are in- 
clined equally toward each other at the top, forming an angle of 45°. Find, 
by arithmetic, the length of the ladders? Ans. 26.13 ft. 

Union Co. Ex. List. 

78. Two trees stand on opposite sides of a stream 120 feet wide; theheight 
of one tree is to the width of the stream as 5 is to 4, and the width of the 
stream is to the height of the other as 4 is to 5; what is the distance between 
their tops? Ans. 136.95 ft. 

79. How many gallons of water will fill a circular cistern 6 feet deep 
and 4 feet in diameter? Ans. 564.0162 gal. 

80. A cube of silver, whose diagonal is 6 inches, was evenly plated 
with gold; if 4 cubic inches of gold were used, how thick was the plating? 

Ans. 6is- in - 

81. Required the distance between the lower corner and the opposite 
upper corner of a room 60 feet long, 32 feet wide, and 51 feet high? 

Ans. 85 ft. 

82. How deep must be a rectangular box whose base inside is 4 inches 
by 4 inches to hold a quart, dry measure? Ans. 4.2 cu. in. 

83. A fly is in the center of the floor of a room 30 feet long, 20 feet 
wide, and 12 feet high. How far will it travel by the shortest path to one 
of the upper corners of the ceiling? Ans. 24.67-f-ft. 

84. A corn crib 25 feet long holds 125 bushels. How many bushels will 
one of like shape and 35 feet long hold ? 

85. Let a cube be inscribed in a sphere, a second sphere in this cube, a 
second cube in this sphere, and so on; find the diameter of the 7th sphere, 
if thatof the first is 27 inches. (2). What is the volume of all the spheres 
so inscribed including the first? Ans. . 

86. The area of a rectangular building lot is 720 sq. ft.; its sides are as 
4 to 5; what will it cost to excavate the earth 7 feet deep at 36^ per cubic 
yard ? A ns. $76.20. 

87. A owns % and B the remainder of a field 60 rods long and 30 rods 
wide at one end and 20 rods wide at the other end, both ends being parallel 
to the same side of the field They propose to lay out through it, parallel 
with the ends, a road one rod wide leaving A's % of the remainder at the 
wide end and B's % at the narrow end of the field. Required the location 
and area of the road. Ans . 

88. The diameter of a circular field is 240 rods. How much grass will 
be left after 7 horses have eaten all they can reach, the ropes which are al- 
lowed them being of equal lengths and attached to posts so located that each 
can touch his neighbor's territory and none can reach beyond the boundary 
of the field? Ans. 62.831853 A. 

89. What is the diameter of a circle inclosing three equal tangent circles, 
if the area inclosed by the three equal circles is 1 acre? Ans. , 



MENSURATION. 333 

90. What is Ihe diameter of a circle inclosing four equal tangent circles 
each being tangent to the the required circle, if the area inclosed by the 
four equal circles is L acre? A?is. i?— 4 v /[5(4 — ~)]v 7 2-4-l)-r-(4 — ~). 

91. What is Ihe greatest number of stakes that can be driven one foot 
apart on a rectangular lot whose length is 30 feet and width 20 feet? 

92. What is the greatest number of inch balls that can be put in a box 15 
inches long, 9 inches wide, and G inches high? Ans 

93. A conical vessel 6 inches in diameter and 10 inches deep is full of 
water. A heavy ball 8 inches in diameter, is put into the vessel; how much 
water will flow out? Ans . 

94. How far above the surface of the earth would a person have to ascend 
in order that 3*j of its surface would be visible? Ans. 8000 mi. 

95. Where must a frustum of a cone be sawed in two parts, to have 
equal solidities, if the frustum is 10 feet long, 2 feet in diameter at one end, 
and feet at the other? Ans . 

96. At the three corners of a rectangular field 50 feet long and 40 feet 
wide, stands three trees whose heights are 60, 80, and 70 feet. Locate 
the point where a ladder must be placed so that without moving it at the 
base it will touch the tops of the three trees, and find the length of the lad 
der. What must be the height of a tree at the fourth corner so that the 
same ladder will reach the top, the foot of the ladder not being moved? 

97. A horse is tied to a corner of a barn 50 feet long and 30 feet wide; 
what is the area of the surface over which the horse can graze, if the rope 
is 80 feet long? Ans. . 

98 How many cubic feet in a stone 32 feet high, whose lower base is a 
rectangle, 10 feet by 4 feet and the upper base 8 feet by \% feet? 

Ans. 805^3 cu. ft 

99. To what height above the ground would a platform, 10 feet by 6 
feet,*have to be elevated so that 720 sq. ft. of surface would be invisible to a 
man standing at the center of the platform, the man being 5 feet high? 

100. Required the side of the least equilateral triangle that will cir- 
cumscribe seven circles, each 20 inches in diameter. Ans. 89 2S203 in. 

101. Required the sides of the least right triangle that will circumscribe 
seven circles each 20 inches in diameter. Ans 123.9320 in. and 

107.3205 in. 

102. How long a ladder will be required to reach a window 40 feet from 
the ground, if the distance of the foot of the J adder from the wall is § of the 
length of the ladder. Ans. 50 ft. 

103. A circular park is crossed by a straight path cutting off ^£ of the 
circumference; the part cut off contains 10 acres Find the diameter of the 
park. Ans. 150 rd., nearly 

104. Find the length of the minute-hand of a clock, whose extreme point 
moves 5 ft. 5.9736 in., in 1 da. 18 hr. ? A ns. 3 in. 

105. A, B, and C, own a triangular tract of land. Their houses are located 
at the vertices of the triangle; where must they locate a well to be used in 
common so that the distance irom the houses to the well will be the same, 
the distance from A to B being T20 rods, from B to C 90 rods and A to C 
80 rods. Ans. . 

106. A horse is tethered from one corner of an equilateral triangular 
building whose sides are 100 feet, by a rope 175 feet long. Over what area 
can he graze? Ans. 90021.109181 sq. ft. 

107. Find the area of the triangle formed by joining the centers of the 
squares constructed on the sides of an equilateral triangle, whose sides are 
20 feet? Ans. 



334 



FINKEL'S SOLUTION BOOK. 



GEOMETRY 



1. Geometry is 

extension. 



the science that treats of position ana 



Si 



Plane. 
Solid. 




2, Pure Geometry s 9 

Problem. — To bisect a given triangle by a line draw it from 
a random point in one of its sides. 

Demonstration. — Let ABC be the given triangle, D a random 
point in the side BC, and E the middle 
point of BC. Join A and D, A and E. 
Draw EF parallel to AD. Draw DF. 
Then DE bisects the triangle ABC. 
For the triangle ABE is equivalent to 
the triangle AEC (?). The triangle 
AED is equivalent to the triangle FIG. 1. 

ADE (?). Hence, ABDE is equivalent to ABE (?)and, 
therefore, DE bisects the triangle^4i> C. Q. E. D. 

Proposition. — The square described upon the hypotenuse of 
a right triangle is equal to the sum of the squares of the other two 
sides. 

I. Demonstration. — Let CFD be any right triangle, right an- 
gled at F and let A C, CP, and DM 
be the squares described upon its 
sides. Then the square A C is equal 
to the sum of the squares CP and 
DM. Through F, draw QF per- 
pendicular to AB and produce it to 
meet OP produced, in G\ also pro- 
duce BC to meet OP in /and AD 
to meet OP produced, in R. Draw 
GH parallel to PD, and BT par- 
allel to CF. Draw AE. Now the 
triangles COI and DEC are 
equal ( ?). Hence, Cf=CD=CB, 
and therefore the square CP=the 
parallelogram CG (?)=the paral- 
lelogram BE (?)=the rectangle FIG- 2. 
BK (?). In like manner, the square DM can be proved equal 
to the rectangle AK. Hence, the square A C=the square CP-\- 
the square DM. Q. E. D. 




GEOMETRY. 



335 



II. Demonstration. — Let ED Che any right triangle, right 
angled at D. On the sides DE and 
DC construct the squares EDHG and 
DCBM respectively. Produce GE 
and BC until they meet in E, form- 
ing the square E B A G. On EC, 
the hypotenuse, construct the square 
ECKI. Then the square ECKI is 
equal to the sum of the squares EDHG 
and D CBM. For, the square GFBA 
is equal to GEDH+D CBM+ 
2 EDCE (=4ECF). The square 
GFBA is also equal to the square 
ECKI + 4ECF. Hence, ECKI+ FIG 3 

4ECF=± GEDH + DCBM+4ECF (?). Whence, ECKI 
= GEDH+D CBM. Q. E. D. 




Proposition. — In any triangle, 
ing the feet of the perpendiculars is 
larfrom the opposite vertex. 



each angle formed by join- 
bisected by the perpc?idicu- 



Demonstration. — Let ABC be any triangle and AD, BE, and 
CE the three perpendiculars. Join D and E, D and F, and E, 
and F. 



In the right triangles AEB and AFC, the angle BA C is 
common to both. Therefore, they are similar. Hence, AB'.AC 
=AE:AF. Now the triangles BA C and FAE have the angle 
FAE common and the including sides proportional. Therefore, 



J n a 
angle 



they are similar, and the angle AFE=the angle A CB. 

similar manner we may prove that the angle DFB=the 

A CB ; the angle A FB==the angle DEB. 

From this it follows that the angle CAF 

— the angle EFA=the angle CFB — 

the angle DEB. Hence, angle EFC=an- 

gle CFD and the angle FED is bisected 

by the perpendicular C F. In a similar 

manner, it can be proved that A D bisects 

the angle FDE and EB bisects the angle 

ABC. Q.E.D. FIG. 4. 

Problem. — From a given point in an arc less than a semi- 
circumference, draw a chord of the circle which will be bisected 
by the chord of tne given arc. 

Demonstration. — Let ABDC be the given circle, AB the 
given arc, AB the chord of the arc, and P any point of the arc 




336 



FINKEL'S SOLUTION BOOK. 




A PC, Draw the diameter POC and on the radius PO as a di- 
ameter describe the circle PEO. Then 
through the points^, and G, of inter- 
section draw the chords PD and PF 
respectively, and they will be bisected 
at the points E and G. For draw 
DC and OE. Then the triangles 
PEO and PD C are right triangles(?) 
and are also similar (?). Since PEO 
and PD Care similar, the line OE 
is parallel to DC, and since O is the 
middle point of P C,E is the middle 
point of PD( ?). In like manner, G 
is the middle poinc of PF. QE. F. 

Discussion. — There are, in general, two solutions. When arc 
AB is diminished until B coincides with A, there is no solution. 
When AB is a semi-circumference, there is one solution and the 
chord is the diameter POC. 

Proposition. — If two equal straight lines intersect each 
other anywhere at right angles, the quadrilateral formed by join- 
ing their extremities is equivalent to half the square on either 
straight line. 

Demonstration. — Let AB and CD be two equal straight lines 
intersecting each other at right angles 
at E. Join their extremities, form- 
ing the quadrilateral A CBD. Then 
ACBD is equivalent to half the 
square of AB or CD. For the area 
of the triangle A CB equals \{AB 
X CD) and the area of the triangle 
ADB equals \{ABy^ED). Hence, 
the area of A CBD=±{ABXED) + 
%(ABxEC)=±AB\ED+EC) = 
±(ABxCD). But CD equals AB, 
by hypothesis. Hence, ACBD 
0. E. D. 




-iAB" 



A PROBLEM IN MODERN GEOMETRY. 



An equilateral hyperbola passes through the middle points 
D, E, and F of the sides BC, AC, and AB of the triangle 
ABC, and cutting those sides in order in a, (3, and y. Show 
that the lines Aa, Bfi, and Cy intersect in a point the locus of 
which is the circumscribing circle of the triangle ABC. 

Solution. — The equation to any conic is zta 2 -\-vfi 2 -\-wy 2 -\- 
2u'fiy+2v'ay-\-2w / a/3=0.::. (1). D is (0, \a sin C, ha sinB) ; 
E t (UsinC, 0, UsinA) ; F, (-Jcsin B, %cs'mA,0). These 



\ 

GEOMETRY. 337 

points being on (1), we should have c 2 v-{-b 2 w-{-2bcu'—0 . . . (2), 
c 2 u-{-a 2 w+2acv'=0 (3), b 2 u-\-a 2 v-{-2abw' = (4). 

Whence ^=— (au' — bv' — cw') .... (5), v= — ( bv' — cw' — au') .. 

bc^ j \ / ac \ j 

. . . (6), w=—(cw / — au' — bv') ....(7). Substituting in the con- 
dition u-\-v-\-w — 2u' cosA — 2v' cosi? — 2w'cosC=0 (8) 

that (1) is an equilateral hyperbola, 

a 2 ( au' — bv' — cvj' ) -\-b 2 ( bv' — cw' — au' ) -\-c 2 ( cw' — au' — bv' ) 

abc 
—2u'co*A—2v' cos B— 2w' cos C=0 (9). Clearing of frac- 
tions and noticing that 2abc cosA—a(b 2 -{-c 2 — a 2 ) (10), 

2abc co*B=b(a 2 +c 2 — b 2 ) . . . .(11), 2abc cos C=c(a 2 -\-b 2 —c 2 ) 

(12), and reducing, u' cos A -\-v' cos B-\-iv' cos C=0. . . .(13). 

Substituting (5), (6), and (7) in (1) an clearing of fractions, 
a 2 (au'—bv'—cw')a 2 -{-b 2 (bv'—cw'—au')(3 2 -{-c 2 (cw'—au'—bv') 
+y 2 -\-2u'abc(3y-\-2v'abcay+2iv'abca/3=<d . . . (14). Where this 

6 2 6 

cuts BC, a=0, and (14) gives b 2 (bv'—cw'—au')~-\-2abcu'- 

v ' ' t/2 ' y 

== — c 2 (civ' — an' — bv') . . . (15), whence for the point a ; a 1 =0, 

„ C Cll' bv' au' t-tn\ -r. 

Pi = 7 — / 1 y / ~/Y\ . • . -(16). By symmetry, for the point. 

b — cu -{-bv — au ' l K ' ' ' J l 

c civ' — au' — bv' _ ,". _, 

p,a. ? = — - — - — — v , # 2 =0 .... (17). The equation to 

a — cw -{-au — bv * s v ' l 

Aa is found to be b( — cw'-{- bv' — aw')/3 — (cw' — bv' — au')y=0 . 
... (18) ; to Bfi, a(—cw'-{-au' — bv')a—c(cw' — au'—bv')y=0 . 

(19) ; and to Cy, b(—au'+bv'—cw')/3 

— a(au' — bv' — cw')a=0 .... (20), any two of which meet in 






(Y 



bc( — cw'-\-bv' — au' ) ( cw' — au' — bv' ) 

" a — 

a, ac ( cw' — bv' — au' ) ( — cw'-\-au' — bv' ) 

p= ^ , 

/ ab( — cw'-\-bv' — au')( — cw' -\-au' — bv') 
Y = ~D[ (21 '' 

The circumscribing circle is a fiy + bay-\-ca fi=0 (22), 

which is satisfied by (21) on condition (13), proving the proposi- 
tion. 

Note. — This problem was solved by Professor William Hoover, A. M., 
Ph. D., Professor of Mathematics and Astronomy in the Ohio University, 
Athens, Ohio, who is one the leading mathematicians in the United States, 
and whose biography follows. 



33S FINKEL'S SOLUTION BOOK, 



BIOGRAPHY. 
PROF. WILLIAM HOOVER, A. M., PH. D. 



Professor Hoover was born in the village of Smithville, Wayne county, 
Ohio, October 17, 1850, and is the oldest of a family of seven children. 
Both parents are living in the village where he was born, still enjoying 
good health. 

Up to the age of fifteen he attended the public schools, and for two or 
three years after, a local academy. Owing to needy circumstances he was 
obliged to work for his living quite early, and almost permanently closed 
attendance at any kind of school at eighteen years of age, sometime before 
which, going into a store in the county seat, as clerk. Nothing could have 
been farther from his taste than this work, having been thoroughly in love 
with study and books long before. After spending two or three years in 
this way, he went to teaching, about the year 1869, and he has been regularly 
engaged in his favorite profession to the present day. 

He attended Wittenberg College and Oberlin College one term each, a 
thing having very little bearing on his education. He studied no mathe 
matics at either place, excepting a little descriptive astronomy at the latter. 

After teaching three winters of country school, with indifferent success, 
he was chosen, in 1871, a teacher in the Bellefontaine, Ohio, High School, 
serving one year, when he was given a place in the public schools of South 
Bend, Ind. Remaining there two years, he was invited to return to Beile- 
fontaine as superintendent of schools. He afterwards served in the same 
capacity in Wapakoneta, O., two years, and as principal of the second dis- 
trict school of Dayton, O. In 1883, he was elected professor of mathemat- 
ics and astronomy in the Ohio University, Athens, Ohio, where he is still 
in service. 

Through all his career of teaching, Professor Hoover has been an inces- 
sant student, devoting himself largely to original investigations in mathe- 
matics. Although his pretentions in other lines are very modest, he is em- 
inently proficient in literature, language, and history. Before going into 
college work he had collected a good library. He is indebted to no one for 
any attainments made in the more advanced of these lines, but by indefati- 
gable energy and perseverance he has made himself the cultured, classic, and 
renowned scholar he is. 

He has always been a thorough teacher, aiming to lead pupils to a mas- 
tery of subjects under consideration. His habits of mind and preparation 
for the work show him specially adapted to his present position, where he 
has met great success. He studies methods of teaching mathematics, 
which in the higher parts is supposed to be dry and uninteresting. He sets 
the example of enthusiasm as a teacher, and rarely fails to impress upon 
the minds of his students the immense and varied applications of mathe- 
matics. He is kind and patient in the class-room and is held in the highest 
esteem by his students. He is ever ready to aid the patient student inquir- 
ing after truth. It seems to be a characteristic of eminent mathematicians 
that they desire to help others to the same heights to which they them- 
selves have climbed. This was true of Prof. Seitz; it is true of Dr. Martin; 
and it is true of Prof. Hoover. 

In 1879, Wooster University conferred upon Prof. Hoover the degree of 
Master of Arts, and, in 1886, the degree of Doctor of Philosophy cum laude, 
he submitting a thesis on Cometary Perturbations. In 1889, he was 
elected a member of the London Mathematical Society and is the only 
man in his state enjoying this honor. In 1890, he was elected a member of 
th& New York Mathematical Society. He has been a member of the Asso- 



\ 




^-/cn^^e^c/ c 



GEOMETRY. 339 

ciation for the Advancement of Science for several years. Papers accepted 
by the association at the meetings at Cleveland, Ohio, and at Washington, 
D. C, have been presented on "The Preliminary Orbit of the Ninth Comet 
of 1886," and ''On the Mean Logarithmic Distance of Pairs of Points in 
Two Intersecting Lines." He is in charge of the correspondence work in 
mathematics in the Chautauqua College of Liberal Arts and of the mathe- 
matical classes in the summer school at Lake Chautauqua, the principal of 
which is the distinguished Dr. William R. Harper, president of the new 
Chicago University. The selection of Professor Hoover for this latter po- 
sition is of the greatest credit, as his work is brought into comparison with 
some of the best done anywhere. 

He is a critical readei and student of the best American and European 
writers, and besides, is a frequent contributor to various mathematical jour- 
nals, the principal of which are School Visitor, Mathematical Messenger, 
Mathematical Magazine, Mathematical Visitor, Analyst, Annals of Math- 
ematics, and Educational Times, of London, England. 

His style is concise and his aim is elegance in form of expression of 
mathematical thought. While greatly interested in the various branches of 
pure mathematics, he is specially interested in the applications to the ad- 
vanced departments of Astronomy, Mechanics, and the Physical Sciences 
— such as Heat, Optics, Electricity, and Magnetism. The "electives" of- 
fered in the advanced work for students in his University are among the 
best mathematics pursued any where in this country. 

He is an active member of the Presbyterian church and greatly interested 
in everv branch of church work. He has been an elder for a number of 
years and was chosen a delegate to the General Assembly meeting at Port- 
land, Oregon, in May, 1892, serving the church in this capacity with fidel- 
ity and intelligence. In this biography of Professor Hoover, there is a val- 
uable lesson to be learned. It is this: Energy and perseverance will bring 
a sure reward to earnest effort. We see how the clerk in a county seat 
store, in embarrassing circumstances and unknown to the world of thinkers, 
became the well known Professor of Mathematics and Astronomy in one of 
the leading Institutions of learning in the State of Ohio. "Not to know 
him argues yourself unknown." 



THE NINE-POINT CIRCLE. 

Proposition. — If a circle be .described about the fed al trian- 
gle of any triangle, it will pass tli rough ilic middle points of the lines 
drawn from the orthocentcr to the vertices of the triangle, and 
through the middle points of the sides of the triangle, in all, through 
nine points. (2) The center of the nine-poi??t circle is the mid- 
dle point of the line joining the orthocentcr and the center of the 
circ?imscribing circle of the triangle. (5) The radius of the nine - 
point circle is half the radius of the circumscribing circle of the 
triangle. (4) The centroid of the triangle also lies on the line 
joining the orthocentcr and the center of the circu?nscribing circle 
of the triangle , and divides it in the ratio of 2\1. (5) The sides of 
the pedal triangle intersect the sides of the given triangle in the 
radical axis of the circumscribing and nine-point circles. (5) The 
nine-point circ 1 touches the inscribed and escribed circles of the 
triangle. 



340 FINKEL'S SOLUTION BOOK. 

The Pedal Triangle is a triangle formed by joining the 
feet of the perpendiculars drawn from the vertices of a triangle 
to the opposite sides. 

The Orthoceflter is the point of intersection of these per- 
pendiculars. 

Medial Lines, or Medians, are lines drawn from the 
vertices of a triangle to the middle point of the opposite sides. 

The Centroid is the point of intersection of the medians. 

The Madical Axis of two circles is the locus of the points 
whose powers with respect to the two circles are equal. 

Demonstration. — Let AB C be any triangle, AD, BE, and 
CE the perpendiculars from the vertices to the opposite sides of 
the triangle. O is the orthocenter. Join the points F, E, and 
D. Then FED is the pedal triangle. About this triangle, de- 
scribe the circle FEHDK. It will then pass through the mid- 
dle points Z, N, and R of the lines, OA, OF, and OC, and the 
middle points H, G, and F of the sides AF, BC, and A C, in all, 
through nine points. 

Since the angles AFO and AEO of the quadrilateral are both 
right angles a circle may be described about it. For the same 
reason circles may be described about the quadrilaterals EFDO 
and OD CF. Draw the lines FR and R G. Now the angles 
ERE and FDE are equal, being measured by half the same 
arc FE. But FDE equals 2EDE, because AD bisects the 
angle FDE. .'. FRO equals 2FDB. Both being measured by 
the same arc OF*, and FRO being two times FDL, FRO is an 
angle at the center; therefore, since OC is the diameter of the 
circle circumscribing FODC, R is the middle point of OC. In 
like manner it may be proved that OB and OA are bisected in 
the points N and L respectively. Draw the line RG. The 
angles RGC and RGB are equal to two right angles. Also the 
angles R GB and RED are equal to two right angles, because 
they are opposite angles of a quadrilateral inscribed in a circle. 
Therefore RGC is equal to RED, But RED is equal to 
OBD, because both are measured by half the arc OD. .-. The 
angle RGC equals the angle OBD, and consequently the line 
RG is parallel to the line OB. But, since RG bisects OC in R 
and is parallel to OB, it bisects B C in G. In like manner, it may 
be shown that AB and A C are bisected by the nine-point circle 
in the points //and K respectively. Hence, the circle passes, in 
all, through nine points. Q. E. D. 

(2.) Draw the perpendiculars GF, KP, and HP from the 
middle points of the sides of the triangle. They all meet in a 
common point P which is the center of the circumscribing circle 
of the triangle. With P as a center and radius equal to PB, 



GEOMETRY. 



341 



describe the circumscribing circle. Draw the perpendiculars SY 
SJ, and SZ to the middle points of the chords FK, EH, and 




FIG. 6. 
DG. These all meet in the same point S, which is the center of 
the nine-point circle. In the trapezoid PHEO, since SJ bi- 
sects EH m J and is parallel to EH, it bisects OP in S. Hence, 
the center of the nine-point circle is the middle point of the line 
joining the orthocenter and center of the circumscribing circle. 

Q. E. B. 

(3.) Draw the lines KN and PB. Since the angle KFN \s 
a right angle, the line KN is a diameter of the nine-point cir- 
cle. KP=\BO=BN. Since KP and BN are equal and 
parallel, KPBN is a parallelogram, and consequently KJV—BP. 
.-. SN==^£P. But SN is the radius of the nine-point circle 
and BP is the radius of the circumscribing circle of the triangle. 
Hence, the radius of the nine-point circle is half the radius of 
the circumscribing circle. Q. E. D. 

(4.) Draw the medial lines BK, AG, and CH. Draw the 
line KH. Now the triangles KPH and BOC are similar, be- 



342 



FINKEL'S SOLUTION BOOK. 



cause the sides of the one are respectively parallel to the sides o 
the other, and the line UK\s half the line B C, because H and 
^Tare the middle points of the sides AB and A C. .-. BO=2KP. 
The triangles KPQ and BOQ are similar, because the angles of 
one are respectively equal to the angles of the other. Then we have 
KP;KQ\\BO\BQ or KP'.BO .:KQ. BQ. But B0=2KP. 
.\ BQ—2ICQ. .'. Q is the centroid and divides the line joining 
orthocenter and the center of the circumscribing circle in the 
ratio of2:l. Q.E. D. 

Hence the line joining the centers of the circumscribing and 
nine-point circles is divided harmonically in the ratio of 2 : 1 by 
the centroid and orthocenter of the triangles, These two points 
are therefore centers of similitude of the circumscribing and 
nine-point circles. .\ Any line drawn through either of these 
points is divided by the circumferences in the ratio of 2 : 1. 

(5. ) Produce FE till it meets BC in P / . Since two opposite an- 
gles of the quadrilateral BE PC are equal to two right angles, a 
circle may be circumscribed about it. Then we have P / E. P / F 
==P / B. P / C ; therefore the tangents from P / to the circles are 
equal. Q. E. D. 

(6. ) Let O be the orthocenter, and / and Q the centers of the in- 
scribed and circumscribed circles. Produce AI to bisect the arc 
BC in T. Bisect AO in L, and 
join GZ, cutting A Pin S. The 
nine-point circle passes through 
G, /?, and B, and D is a right an- 
gle. Hence, GK is a diameter, 
and is therefore —R=QA. There- 
fore GL and QA are parallel. But 
QA=QP, therefore GS=GT= 
C7^sin^=2i?sin 2 ^. Also ST 
=2GScosO, 6 being the angle GST 
= GTS. JV being the center of 
the nine-point circle, its radius 
=JVG=^B ; and r being the radius 
of the inscribed circle, it is required 
to show that NI=NG—r. NI 2 
=SW 2 +SI t >—2SlV.S/cosd. Sub- 
stitute SM=%jR—GS; SI=TI— FIG. 7. 
ST=2BsiniA—2GScosO; and GS=2Psin 2 iA, to prove the 
proposition. If J be the center of the escribed circle touch 
ing B C, r x its radius, it is shown in a similar way that 
m=NG\r v 




GEOMETRY. 843 

PROPOSITIONS. 

1. The lines which join the middle points of adjacent sides of any quad- 
rilateral, form a parallelogram. 

2. Two medians of a triangle are equal; prove* (without assuming that 
they trisect each other) that the triangle is isoscles. 

3. In an indefinite straight line AB find a point equally distant from 
two given points which are not both on AB. 

When does this problem not admit of solution? 
Construct a right triangle having given: 

4. The hypotenuse and the difference of the sides. 

5. The perimeter and an acute angle. 

6. The difference of the sides and an acute angle. 

7. Construct a triangle having given the medians. 

8. Construct a triangle, having given the base, the vertical angle, and(l) 
the sum or (2) the difference of the sides. 

9. Describe a circle which shall touch a given circle at a given point, and 
also touch a given straight line. 

10. Describe a circle which shall pass through two given points and 
be tangent to a given line. 

11. Find the point inside a given triangle at which the sides subtend 
equal angles. 

12. Describe a circle which shall be tangent to two intersecting straight 
lines and passing through a given point. 

13. Divide a triangle in two equal parts by a line perpendicular to a 
side. 

14. Inscribe in a triangle, a rectangle similar to a given rectangle. 

15. Construct an equilateral triangle equivalent to a given square. 

16. Trisect a triangle by straight lines drawn from a given point in one 
of its sides. 

17. Draw through a given point a straight line, so that the part of it in- 
tercepted between a given straight line and a given circle maybe divided 
at the given point in a given ratio. 

IS. Construct a circle equivalent to the sum of three given circles. 

19. Find the locus of a point such that the sum of its distances from three 
given planes is equal to a given straight line. 

20. Construct a sphere tangent to three given spheres and passing 
througn a given point. 

21. Draw a circle tangent to three given circles. 

Note. — This proposition is known as the Taction Problem. It was pro- 
posed and solved by Apollonius, of Pergse, A. D. 200. His solution was indi- 
rect, reducing the problem to ever simpler and simpler problems. It was 
lost for centuries, but was restored by Vieta. The first direct solution was 
given by Gergonne, 1813. An elegant solution to this problem is given by 
Prof. E. B. Seitz, School Visitor, Vol. IV, p. 61. 

22. Construct a sphere tangent to four given spheres. 
Note. — This problem was first solved by Fermat (1001 — 1065). 

23. The perpendicular from the center of gravity of a tetrahedron to any 
plane without the tetrahedron is one-fourth of the sum of the perpendic- 
ulars from the vertices to the same plane. 



344 



FINKEL'S SOLUTION BOOK. 



PROBABILITY PROBLEMS. 



I. If three pennies be piled up at random on a horizontal 
plane, what is the probability that the pile will not fall down? 

The pile will stand if the common center of gravity of the sec- 
ond and third coins falls on the surface of the first or bottom 
coin. 

Let r be the radius of a penny ; then the center of the second 
coin may fall anywhere in a cir- 
cle whose radius is 2r and cen- 
ter the center of the surface* of 
the first or bottom coin, and the 
center of the third coin may fall 
anywhere in a circle whose ra- 
dius is 2r and the center the 
center of the surface of the sec- 
ond coin. The number of posi- 
tions of the center of the second 
coin is therefore proportional to 
47Tr 2 , and for every one of these 
positions the center of the third 
coin can have 47rr 2 ; hence the 
total number of positions of the 
second and third coins is proportional to 167T 2 /' 4 . 

We must now determine in how many of these 167T 2 r 4 posi- 
tions the pile will stand. 

Let A be the center of the first or bottom coin, and B the cen- 
ter of the second coin. Take AD=AB, and with center D 
and radius 2r describe the arc CHJ. If the center of the third 
coin is on the surface CFJH, the second and third coins will re- 
main on the first, since BN=NH, B T= TC, and the pile will 
not fall down. 

When AB is not greater than -Jr, the circle CHJ will not cut 
the surface of the second coin, and the pile will stand if the cen- 
ter of the third coin is anywhere on the second. 

Let ^4.Z>=^Z>=#,.S=surface CFJH, and /=the probability 

required; then DB=2x,BG=- — — , DG=- 




4x 



4x 



CG= i[ 16 ^- (5r2 - 4 ^ 2)2 0^ arc CI==rcOBrl Q r \rt** ^ 



— r^cos' 



16 



PROBABILITY PROBLEMS. 345 

r 8r2 4 7^* 2 )— |[l6r 4 -(5r 2 — 4* 2 ) 2 T, and 

1 /*^ r 1 /* r 

i>=— — — - - / 7tr 2 .27rxdx-\-— - — - — -/ S.lnxdx, 

— 1 / (S — 7tr 2 )xdx. I r 2 cos -1 ! - — — Jxdx, 

=2r x cos v-8^-y +ir (OS v~4^~J 

_l r 2r i6r 4_( 5r 2_ 4x 2)2T' 2 j T^lGr 4 — (5r 2 — 4aT 2 ) 2 "| ^X 

^^cos" 1 ^— ^"-)— 4( 5r2_ 4* 2 )[^16r4—(5r 2 — 4* 2 ) 2 ~|* 

1 16 87rr 4 L T V 4r* y 

(3r*+4* 2 x . y5r 3 -4r\ , 1 /r . .. 

V l6 ^_(5,»-4^)»]' xr =l- i | i (^i5_2sin-i). 

Note. — This solution is due to Artemas Martin, M. A., Ph. D., LL. D., 
member of the London Mathematical Society, member of the Edinburg 
Mathematical Society, member of the Mathematical Society of France, 
member of the New York Mathematical Society, member of the Philo- 
sophical Society of Washington and Fellow of the American Association 
for the Advancement of Science, Washington, D. C, who is one of the 
great peers of mathematical science. 



346 FINKEL'S SOLUTION BOOK. 



BIOGRAPHY. 
ARTEMAS MARTIN, M- A., PH. D., LL. D. 



This eminent mathematician was born in Steuben county, N. Y., Au- 
gust 3, 1835. Early, his parents moved to Venango county, Pa., where they 
lived for many years. Dr. Martin had no schooling in his early boyhood, 
except a little primary instruction; but by self-application and indefatiga- 
ble energy which have told the story of many a great man, he has become 
familiar to; every mathematician and lover of science in every civilized 
country of the world. 

He was never a pupil at school, except when quite small, until in his 
fourteenth year. He had learned to read and write at home, but knew 
nothing of Arithmetic. At fourteen he commenced the study of Arithme- 
tic, and after spending two winters in the district school, he commenced 
the study of Algebra. At seventeen, he studied Algebra, Geometry, Nat- 
ural Philosophy, and Chemistry in the Franklin Select School, walking 
two and one-half miles night and morning. Three years after, he spent 
two and one-half months in the Franklin Academy, studying Algebra and 
Trigonometry. This finished his schooling. He taught district schools 
four winters, but not in succession. He was raised on a farm, and worked 
at farming and gardening in the summer; chopped wood in the winter; 
and after the discovery of oil in Venango county, worked at drilling oil 
wells a part of his time, always devoting his "spare moments" to study. 

In the spring of 1869,$the family moved to Erie county, Pa., where he re- 
sided until he entered the U. S. Coast Survey Office in 1885. While in 
Erie county, after 1871, he was engaged in market-gardening, which he 
carried on with great care and skill. He began his mathematical career 
when in his eighteenth year, by contributing solutions to the Pittsburg 
Almanac, soon after contributing problems to the "Riddler Column" of the 
Philadelphia Saturday Evening Post, and was one of the leading contribu- 
tors for twenty years. 

In the summer of ,1864 he commenced contributing problems and solu- 
tions to Clark's School Visitor, afterward the Schoolday Magazine, pub- 
lished in Philadelphia. In June, 1870, he took charge of the "Stairway De- 
partment" as editor, the mathematical department of which he had con- 
ducted for some years before. He continued in charge as mathematial edi- 
tor till the magazine was sold to Scribner & Co., in the spring of 1875, at 
which time it was merged into u St. Nicholas ." 

In September, 1875, he was chosen editor of a department of higher 
mathematics in the Normal Monthly, published by Prof. Edward Brooks, 
Millersville, Pa., and held that position till the Monthly was discontinued 
in August, 1876. He published in the Normal Monthly a series of sixteen 
articles on the Diophantine Analysis. 

In June, 1877, Yale College conferred on him the honorary degree of 
Master of Arts (M. A.) In April, 1878, he was elected member of the Lon- 
don Mathematical Society. In June, 1882, Rutgers College conferred on 
him the honorary degree of Doctor of Philosophy (Ph. D.) March 7, 1884, 
he was elected a member of the Mathematical Society of France. In April, 
1885, he was elected a member of the Edinburgh Mathematical Society. 
June 10, 1885, Hillsdale College conferred on him the honorary degree of 
Doctor of Laws (LL. D.) February 27, 1886, he was elected a member of 
the Philosophical Society of Washington. In June, 1881, he was elected 
Professor of Mathematics of the Normal School at Warrensburg, Mo., but 
did not accept the position. November 14, 1885, Dr. Martin was appointed 



PROBABILITY PROBLLMS. 347 

Librarian in the office of the U. S. Coast and Geodetic Survey. On August 
26, 1890, he was elected a Fellow of the American Association for the Ad- 
vancement of Science. On April 3, 1891 he was elected a member of the 
New York Mathematical Society. 

All these honors have been worthily bestowed and the Colleges and So- 
cieties conferring them have done honor to themselves in recognizing the 
merits of one who has become such a power in the scientific world through 
his own efforts. 

He has contributed fine problems and solutions to the following journals 
of the United States: School Visitor, A?ialyst, Annals of Mathematics, 
Mathematical Monthly, Illinois Teacher, lovja Instructor, National Edu- 
cator, Yates County Chronicle, Barnes'' Educational Monthly, Wittenberger, 
Maine Farmers' 1 Almanac, Mathematical Messenger, and Educational 
Notes and Queries. Besides other contributions, he contributed thirteen 
articles on "Average" to the Mathematical Department of the Witten- 
berger, edited by Prof. William Hoover. These are believed to be the first 
articles published on that subject in America. 

Dr. Martin has also contributed to the following English mathematical 
periodicals: Lady's, and Gentleman' 's Diary, Messenser o-P Mathematics, 
and The Educational Times a?id Repri?it. 

The Reprint contains a large number of his solutions to difficult "Aver- 
age" and "Probability" problems, which are master-pieces of mathematical 
thought and skill, and they will be lasting monuments to his memory. His 
style is direct, clear and elegant. His solutions are neat, accurate and sim- 
ple. He has that rare faculty of presenting his solution in the simplest 
mathematical language, so that those who have mastered the elements of 
the various branches of mathematics, are able to understand his reasoning. 

Dr. Martin is now (1893) editor of the Mathematical Messenger, and 
The Mathematical Visitor, two of the best mathematical periodicals pub- 
lished in America. These are handsomely arranged and profusely illustra- 
ted with very beautiful diagrams to the solutions, he doing the typesetting 
with his own hand. The typographical work of these journals is said to be 
the finest in America. The best mathematicians from all over the world 
contribute to these two journals. The Mathematical Visitor is devoted to 
Higher Mathematics, while The Mathematical Magazine is devoted to the 
solutions of problems of a more elementary nature. All solutions sent to 
Dr. Martin receive due credit, and if it is possible to find room for them the 
solutions are all published. He has thus encouraged many young aspirants 
to higher fields of mental activity. He is always ready to aid any one who 
is laboring to bring success with his work. He is of a kind and noble dis- 
position and his generous nature is in full sympathy with every diligent 
student who is rising to planes of honor and distinction by self application 
and against adverse circumstances. 

Dr. Martin has a large and valuable mathematical library containing 
many rare and interesting works. His collection of American arithmetics 
and algebras is one of the largest private collections of the kind in this 
country. 



34S FINKEL'S SOLUTION BOOK. 

I. Find the average or mean distance of every point of a square 
from one corner. 

Taking the corner from which the mean distance is to be 
found for the origin of orthogonal co-ordinates, and one of the 
sides of the square for the axis of abscissa, we have for the ele- 
ment of the surface dx dy, and since this element is at a distance 

V( x2J ry 2 ) from the origin, the average — / / dxdy^(x 2 -\-y 2 ) 

= A \ a f a dx^( a ^ +x 2 ) + f a x ^ dx log / + ^ (g2+Jf2) I . But 
2>a z ( Jo Jo x ) 

r 2 j i tf+V(« 2 +* 2 ) , 31 tf+V(tf 2 +* 2 . . r x*dx 

-^ 3 log f +V( ^ + ^ 2) +i^(^ 2 +^ 2 )-^ 3 log e ^+V(^ 2 +^ 2 )j 

.-. Average=^[V2+log e (l+V2)]. 

Note. — This solution is by Prof. J. W. F. Sheffer, Hagerstown, Md., 
whose name may be found attached to the solutions of many difficult prob- 
lems proposed in the leading mathematical journals of the United States. 
The above solution is taken from the Mathematical Messenger, published 
by G. H. Harville, Simsboro, La. 

I. All that is known concerning the veracities of two witnesses, 
A and B, is that B's statements are twice as reliable as A's. 
What is the probability of the truth of the concurrent testimony 
of these two witnesses? 

Let *=the probability of the truth of any one of A's state- 
ments ; then 2*— the probability of any one of B's statements. 
The event did occur if both witnesses tell the truth, the proba- 
bility ot which is *X2*=2* 2 . The event did not occur if both 
testify falsely, the probability of which is (1 — *)X(1 — 2*)=1 
— 3x-\-2x 2 . Hence, the probability of the occurrence of the event 
on the supposition that x is known is 

■P 2* 2 +(l— *)(1— 2*)' 

Now, as the veracity of B can not exceed unity, the greatest 
value of* is found by putting 2*=1, which gives *=-J; hence, 
x can have any value from to \. 

Therefore, the probability in the problem is 

/** , , r* y a r y * x2dx 

Ho P dX ^Jo dX== Vo 2* 2 +(l-*H 1=2*7 
f* x 2 dx 

Jo (8*-3) 2 +7* 
Let 8* — 3=y. Then x=\(y | 3), ex— \dy\ the limits of y are 
1 and — 3, and 

P ~sL 3 yi+7 8.7-3 \-^y2+1 + y*+7j ' f 



PROBABILITY PROBLEMS. 



349 



oppo 



=4-110^2+^ tan"V7. 

Note. — This solution is taken from the Mathematical Magazine, Vol. 
II, p. 122. The solution there given is credited to the author, Prof. William 
Hoover, and Prof. P. H. Philbrick. 

I. A cube is thrown into the air and a random shot fired 
through it; find the chance that shot passed through 
site faces. 

Let AH he the cube. Through P , a point in the face EFGH, 
draw MK parallel to HE, and draw PN perpendicular to HE. 
Now if PA represents the direction of 
the shot, it will pass through the face 
ABCjD, if it strikes the face EFGH 
anywhere within HMPN. 

Y,etAB=\, LKAF=V, lPAK=cf), 
and area HMPJV=u. Then AK=sec6, 
PK=secdtan0, EK=tand , AP= 
sec0sec0, PM=1 — sec tan0, PN=l 
— tan#, u=(l — secfltan0)(l— tanfl ),the 
area of the projection of HMPN on a 
plane perpendicular to PA—ucosdcos(p, 
and that of EEGH=cosdcos(f). 

Since we are to consider all possible FIG- 9. 

directions of the shot with respect to the cube, the points of in- 
tersection of PA with the surface of a sphere whose center is A, 
and radius unity, must be uniformly distributed. An element 
of the surface of this sphere is coscpddd(p. By reason of the 
symmetry of the cube, the required chance is obtained by finding 
the number of ways the shot can pass through the opposite faces 
EEGHand ABCD between the limits 6=0, and B=\n, and 
<p=0 and 0=tan _1 (cos6')=0 / , and the number of ways it can 
pass through the face EEGHbetween the limits 6=0 and 6=\n, 
and cf>=0 and <fr=^ir ; and then dividing the former by the latter. 
Hence, the chance required is 




/Mir /"*' 
/ ucos8cos 2 (f)d p d(f) 
Jo 

?~' /-Kit rV^ ~ 



4 rViK /**' 

TtJo Jo 



cos6 cos 2 (pdddcp, 



J I cosD'cos 2 (f)d 6 d<p 
Jo Jo 

2 rV** 
=~ I ( cos6— s'm8)tan~ 1 (cosO )d8, 

^["(sinfl+cosfljtan-^cosfl)— 0+^2 tan"^^ tan/9) 

-i log.( 1+cos 2 )~| * = -W tan-* (iv/2)+log,(f )-tt]. 

Note. — This solution is due to Professor Enoch Beery Seitz, member 
of the London Mathematical Society, and late Professor of Mathematics 
in the North Missouri State Normal School, Kirksville, Mo. 



350 FINKEL'S SOLUTION BOOK. 



BIOGRAPHY. 
PROF. E. B. SEITZ, M. L. M. S. 



Professor Seitz, the most distinguished mathematician of his day, was 
born in Fairfield Co., O., Aug. 24, 1846, and died at Kirksville, Mo., Oct. 8, 
1883. His father, Daniel Seitz was born in Rockingham Co., Va., Dec. 17, 
1791, and was twice married. His first wife's name was Elizabeth Hite, of 
Fairfield Co., 0.,by whom he had eleven children. His second wife's name was 
Catharine Beery, born in the same county, Apr. 11, 1808, whom he married 
Apr. 15, 1832. This woman was blessed by four sons and three daughters. 
Mr. Seitz followed the occupation of a farmer and was an industrious and 
substantial citizen. He died near Lancaster, O., Oct. 14, 1864, in his seventy- 
third year; having been a resident of Fairfield Co. for sixty-three years. 

Professor Seitz, the third son by his father's second marriage, passed 
his boyhood on a farm, and like most men who have become noted, had 
only the advantages of a common school education. Possessing a great thirst 
for learning, he applied himself diligently to his books in private and be- 
came a very fine scholar in the English branches, especially excelling in 
arithmetic. It was told the author, by his nephew, Mr. Huddle, that when 
Professor Seitz was in the field with a team, he would solve problems while 
the horses rested. Often he would go to the house and get in the garret 
where he had a few algebras upon which he would satiate his intellectual 
appetite. This was very annoying to his father who did not see the future 
greatness of his son, and many and severe were the floggings he receiv- 
ed for going to his favorite retreat to gain a victory over some difficult 
problem upon which he had been studying while following the plow. Though 
the way seemed obstructed, he completed algebra at the age of fifteen, with- 
out an instructor. He chose teaching as his profession which he followed 
with gratifying success until his death. He took a mathematical course in 
the Ohio Wesleyan University in 1870. In 1872, he was elected one of the 
teachers in the Greenville High School, which position he held till 1879. 
On the 24th of June, 1875, he married Miss Anna E. Kerlin, one of Darke 
county's most refined ladies. In 1879, he was elected to the chair of mathe- 
matics in the Missouri State Normal School, Kirksville, Mo., which position 
he held till death called him from the confines of earth, ere his star of fame 
had reached the zenith of its glory. He was stricken by that "demon of 
death," typhoid fever, and passed the mysterious shades, to be numbered 
with the silent majority, on the 8th of October, 1883. On the 11th of March, 
1880, he was elected a member of the London Mathematical Society, being 
the fifth American so honored. 

Professor Seitz was in mathematics what Demosthenes was in oratory; 
Shakespear in poetry; and Napoleon in war: the equal of the best, the peer 
of all the rest. 

He began his mathematical career in 1872, by contributing solutions to 
the problems proposed in the "Stairway" department of the Schoolday 
Magazine, conducted by Artemas Martin. His masterly and original solu- 
tions to difficult Average and Probability problems, soonattracted universal 
attention among mathematicians. Dr. Martin being desirous to know what 
works he had treating on that difficult subject, was greatly surprised to learn 
that he had no works upon the subject, but had learned what he 'knew about 
that difficult department of mathematical science by studying the problems 
and solutions in the Schoolday Magazine Afterwards, he contributed to the 
Analyst, th.e Mathematical Visitor, the Wathematical Magazine, the School 
Visitor, and the Educational Times, of London, Eng. 

In all of these journals, Professor Seitz was second to none, as his logical 
and classical solutions to Average and Probability problems,risingas so many 



w 



'^ISB^' 1 "' 1 *"'' 




1 



.Ai, 



<J2^v~t^V~ 



0-Uy^^3, 




I 



PROBABILITY PROBLEMS. 351 

monuments to his untiring patience and indomitable energy and persever- 
ence will attest. His name first appeared as a contributor to the Educa- 
tional Times in Vol. XVIII., of Reprint, 1873. From this time until his 
death the Reprint is adorned with some of the finest product of his mighty 
intellect. 

On page 21, Vol. II., he has given the above solution. This problem had 
been proposed in 1864 by the great English mathematician, Prof. Wool- 
house, who solved it with great labor. It was said by an eminent mathema- 
tician of that day that the task of writing out a copy of that solution was 
worth more than the book in which it was published. 

No other mathematician seemed to have the courage to investigate the 
problem after Prof. Woolhouse gave his solution to the world, till Professor 
Seitz took it up and demonstrated it so elegantly in half a page of ordinary 
type, that he fairly astonished the mathematicians of both Europe and 
America. Prof. Woolhouse was the best English authority on Probabilities, 
even before Professor Seitz was born. 

It was the solution of this problem that won for Professor Seitz the ac- 
knowledgement of his superior ability over any other man in the world. 

In studying his solutions, one is struck with the simplicity to which he 
has reduced the solutions of some of the most intricate problems When he 
had grasped a problem in its entirity, he had mastered all problems of that 
class. He would so vary the conditions in thinking of one special problem 
and in effecting a solution that he had generalized all similar cases, so ex- 
haustive was his analyses. Behind the words he saw all the ideas repre- 
sented These he translated into symbols, and then he handled the sym- 
bols with a facility that has never been surpassed. 

What he might have accomplished in his maturer years no man may say; 
but at the age of thirty-seven he laid down his pen and gave to God from 
whence it came, the casement and the key of his mighty intellect, leaving his 
impress indelibly stamped upon the thinking and scientific world for all time. 

He was a man of the most singularly blameless life; his disposition was 
amiable; his manner gentle and unobtrusive; and his decision, when cir- 
cumstances demanded it, was prompt and firm as the rocks. He did noth- 
ing from impulse; he carefully considered his course and came to conclu- 
sions which his conscience approved; and when his decision was made, it 
was unalterable. 

Professor Seitz was not only a mathematician, but he was eminently pro- 
ficient in other branches of knowledge. His mind was cast in a gigantic 
mold. "Being devout in heart as well as great in intellect, 'signs and quan- 
tities were to him but symbols of God's eternal truth' and 'he looked up 
through nature up to nature's God.' Professor Seitz, in the very appropriate 
words of Dr. Peabody regarding Benjamin Pierce, Professor of Mathematics 
and Astronomy in Harvard University, 'saw things precisely as they are 
seen by the infinite mind. He held the scales and compasses with which 
eternal wisdom built the earth, and meted out the heavens. As a mathemati- 
cian, he was adored with awe. As a man, he was a christian in the whole 
aim and tenor of life.' " No mathematician was so universally loved and 
honored by his contemporaries as Professor Seitz. 

Professor Seitz did not gain his knowledge from books, for his library 
consisted of only a few books and periodicals. He gained such a profound 
insight in the subtle relations of numbers by close application with which 
he was particularly gifted. He was not a mathematical genius, that is, as 
usually understood, one who is born with mathematical powers fully de- 
veloped. But he was a genius in that he was especially gifted with the 
power to concentrate his mind upon any subject he wished to investigate. 
This happy faculty of concentrating all his powers of mind upon one topic to 
the exclusion of all others, and viewing it from all sides, enabled him to 
proceed with certainty where others would become confused and disheart- 
ened. Thread by thread and step by step, he took up and followed out long 
lines of thought and arrived at correct conclusions. The darker and more 



352 



FINKEL'S SOLUTION BOOK. 



subtle the question appeared to the average mind, the more eagerly he in- 
vestigated it. No conditions were so complicated as to discourage him. 
His logic was overwhelming. 

He left a wife and four sons — one of whom has gone to join his father in 
the realms of eternal peace. His mother, now (1893) eighty-five years old, 
is still living and enjoying good health. 



Table I. — Functions of n and e. 



n = 3-1415926 
?r 2 = 9-8696044 
^8=31-0062761 
s)n= 1-7724539 
log 10 7r= 1-4971499 
log, n= 0-6679358 



;r-i= -3183099 

7r- 2 = -1013212 

n-*= -0322515 

200" -^=63" -6619772 

180 o H-7T=57°-2957795 

=206264"-8 



e =2-71828183 
*a. =7-38905611 
*-i =0-3678794 
<r-2 = o-1353353 
\og 10 e =0-43429448 
log, 10 =2-30258509 



Table II. 



No. 


Square root. 


Cube root. 


2 


1-4142136 


1-2599210 


3 


1-7320508 


1-4422496 


4 


f 2-0000000 


1-5874011 


5 


2-2360680 


1-7099759 


6 


2-4494897 


1-8171206 


7 


2-6457513 


1-9129312 


8 


2-8284271 


2-0000000 


9 


3-0000000 


2-0800837 


10 


3-1622777 


2-1544347 


11 


3-3166248 


2-2239801 


12 


3-4641016 


2-2894286 


13 


3-6055513 


2-3513347 


14 


3-7416574 


2-4101422 


15 


3-8729833 


2-4662121 


16 


4-0000000 


2-5198421 


17 


4-1231056 


2-5712816 


18 


4-2426407 


2-6207414 


19 


4-3588989 


2-6684016 


20 


4-4721360 


2-7144177 


21 


4-5825757 


2-7589243 


22 


4-6904158 


2-8020393 


23 


4-7958315 


2-8438670 


24 


4-8989795 


2-8844991 


25 


5-0000000 


2-9240177 


26 


5-0990195 


2-9624960 


27 


5-1961524 


3-0000000 


28 


5-2915026 


3-0365889 


29 


5-3851648 


3-0723168 


30 


5-4772256 


3-1072325 





Table 


III. 


N. 


log 10 N. 


loge N. 


2 


•3010300 


•69314718 


3 


•4771213 


1-09861229 


5 


•6989700 


1-60943791 


7 


•8450980 


1-94591015 


11 


1-0413927 


2-39789527 


13 


1-1139434 


2-56494936 


17 


1-2304489 


2-83321334 


19 


1-2787536 


2-94443898 


23 


1-3617278 


3-13549422 


29 


1-4623980 


3-36729583 


31 


1-4913617 


3-43398720 


37 


1-5682017 


3-61091791 


41 


1-6127839 


3-71357207 


43 


1-6334685 


3-76120012 


47 


1-6720979 


3-85014760 


53 


1-7242759 


3-97029191 


59 


1-7708520 


4-07753744 


61 


1-7853298 


4-11087386 


67 


1-8260748 


4-20469262 


71 


1-8512583 


4-26267988 


73 


1-8633229 


4-29045944 


79 


1-8976271 


4-36944785 


83 


1-9190781 


4-41884061 


89 


1-9493900 


4-48863637 


97 


1-9867717 


4-57471098 


101 


2-0043214 


4-61512052 


103 


2-0128372 


4-63472899 


107 


2-0293838 


4-67282883 


109 


2-0374265 


4-69134788 



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An Academic and Commercial Arithmetic. By Webster 
Wells. Boston : Leach, Shewell, and Sanborn. 

ALGEBRA. 

The Elements of Algebra. By Geo. Lilley, Ph. D., LL. D. 
Boston, New York, Chicago, and Philadelphia : Silver, Bur- 
dett & Co. 

A College Algebra. By G. A. Wentworth, A. M. Boston : 
Ginn & Co. 

A Complete Algebra. By George W. Smith. New. York and 
Cincinnati : American Book Company. 

Slid J on's Complete Algebra. New York and Chicago: Shel- 
don & Co. 

Numbers Universalized, An Advanced Algebra. By David 
M. Sensenig, M. S. New York, Chicago, and Cincinnati : Ameri- 
can Book Company. 



College Algebra. By Edward A. Bowser, LL. D. Boston : 
D. C. Heath & Co. 

College Algebra. By Webster Wells, S. B. Boston, Chica- 
go, and New York: Leach, Shewell & Sanborn. 

A University Algebra. By Edward Olney. New York and 
Chicago: Sheldon & Co. 

The Normal Elementary Algebra. By Edward Brooks, A. 
M., Ph. D. Philadelphia : Christopher Sower Company. 

Algebra for Colleges. By Simon Newcomb. New York and 
Chicago: Henry Holt & Co. 

A Complete Algebra. By A. Schuyler. New York, Chicago, 
and Cincinnati : American Book Company. 

GEOMETRY. 

Elements of Geometry. By Elias Loomis, LL. D. 

Plane and Solid Geometry. By Edward Brooks, A. M., Ph. D. 
Philadelphia: Christopher Sower Company. 

Elementary Geometry and Trigonometry , University Edition. 
By Edward Olney. New York and Chicago: Sheldon & Co. 

A Treatise on Elementary Geometry. By William Chau- 
venet, LL. D. Philadelphia : J. B. Lippincott & Co. 

Chauvenef s Treatise on Elementary Geometry. By W. E. 
Byerlv. Philadelphia: J. B. Lippincott & Co. 

Plane and Solid Geometry. By Seth T. Stewart. New York, 
Chicago, and Cincinnati: American Book Company. 

The Elements of Geometry. By George Bruce Halstead. 
New York : John Wiley & Sons. 

Elements of Geometry. By Webster Wells, S. B. Boston, 
New York, and Chicago: Leach, Shewell & Sanborn. 

Elements of Geometry, By Joseph Bayma, S. J. San Fran- 
cisco: A. Waldteufel. 

The Elements of Plane and Solid Geometry. By Edward A. 
Bowser. Boston : D. C. Heath & Co. 

Elements of Geometry. Bv Simon Newcomb. New York: 
Henry Holt & Co. 

LegendrJs Geo??zetry and Trigono?netry. New York, Cincin- 
nati and Chicago: American Book Company. 

Analytical Geometry. By Chas. Davies, LL. D. New York, 
Cincinnati, and Chicago : American Book Company. 

Introductory Modern Geometry. By William Benjamin Smith. 
A. M., Ph. D. New York : Macmillan & Co. 

Plane and Solid Geometry. By George A. Wentworth. Bos- 
ton: Ginn & Co. 

TRIGONOMETRY. 

Elements of Plane and Spherical Trigonometry. By H.N. 
Robinson, LL. D. New York, Cincinnati, and Chicago: Ameri- 
can Book Company. 

Plane and Spherical Trigonometry and Surveying. By G. 
A. Wentworth, M. A. Boston and Chicago : Ginn & Co, 



Elements of Trigonometry ^ Surveying, and Navigation. By 
Elias Loomis, LL. D. New York : Harper Bros. 

A Treatise on Trigo7tometry. By Edward A. Bowser, LL. D. 
Boston : D. C. Heath & Co. 

ANALYTICAL GEOMETRY. 

Analytical Geometry. , By Geo. H. Howison, M. D. American 
Book Company. 

Analytical Geometry. By Wm. G. Peck, Ph. D. American 
Book Company. 

Analytical Geometry. By G. A. Wentworth, M. A. Boston 
and Chicago: Ginn & Co. 

Elementary Co-ordinate Geometry. By W. B. Smith, Ph. D. 
Boston and Chicago: Ginn & Co. 

Analytical Geometry. By Joseph Bayma, S.J. San Francis- 
co : A. Waldteufel. 

An Elementary Treatise on Analytical Geometry. By Edward 
A. Bowser LL. D. New York: D. Van Nostrand & Co. 

Elements of Analytical Geometry. By Simon Newcomb, 
Ph. D., New York and Chicago: Henry Holt & Co. 

CALCULUS. 

General Geometry and Calculus. By Edward Olney. New 
York and Chicago: Sheldon & Co. 

Differential and Integral Calculus. By George A. Osborne, 
S. B. Boston, New York, and Chicago: Leach, Shewell & San- 
borne. 

Au Elementary Treatise on the Differential and. Integral Cal- 
culus. By Edward A. Bowser, LL. D. New York: D. Van 
Nostrand & Co. 

Elements of the Differential Calculus. By W. E. Byerly. 
Boston: Ginn & Co. 

Elements of the Differential and Integral Calculus. By 
Simon Newcomb. New York and Chicago: Henry Holt & Co. 

Elements of the Differential and Integral Calculus. By J. 
M. Taylor. Boston and Chicago: Ginn & Co. 

Elements of the Differential and Integral Calculus. By C. 
P. Buckingham. Chicago: S. C. Griggs & Co. 

Elements of the Integral Calculus. By W. E. Byerly. Boston 
and Chicago: Ginn & Co. 

Practical Calculus. By W. G. Peck, LL. D. Chicago: A. 
vS. Barnes & Co. 

Differential and Integral Calculus. Bv Ed. H. Courtenay. 
New York, Chicago, and Cincinnati: American Book Company. 

Ray's Elements of the Infinitesimal Calculus. By James G. 
Clark, A. M. New York, Chicago, and Cincinnati: American 
Book Company. 

Differential and Integral Calculus. By H. N. Robinson, LL. 
D. New York, Chicago, and Cincinnati: American Book Com- 
pany. 



MECHANICS. 

An Elenzcntary Treatise on Analytic Mechanics. By Edward 
A. Bowser, LL. D. New York: D. Van Nostrand. 

Teacher's Hand-book of Mechanics. By James Wright, A. M., 
Ph. D. D. Van Nostrand & Co. 

MISCELLANEOUS MATHEMATICAL WORKS. 

The Philosophy of Arithmetic. By Edward Brooks, A. M., 
Ph. D. Philadelphia: Christopher Sower Company. 

The Philosophy of Mathematics. By Albert Bledsoe, LL. D. 
Philadelphia : J. B. Lippincott & Co. 

Mensuration. By George B. Halstead, A. M., Ph. D. Boston 
ond Chicago: Ginn & Co. 

determinants. By Paul H. Hanus, B. S. Boston and Chi- 
cago: Ginn & Co. 

Elements of Quaternions. By B. S. Hardy, Ph. D. Boston 
and Chicago: Ginn & Co. 

MATHEMATICAL JOURNALS. 

The School Visitor. Devoted to Practical Mathematics, Ex- 
amination Work, Notes, Queries and Answers. Edited and 
Published by John S. Royer, Supt. of School, Versailles, O. Pub- 
lished monthly. Price, $1.00 per year. 

One of the most active and vigorous school journals in the State. It has 
rendered uncalculable good to thousands of teachers. Every earnest teacher 
should take this Journal. 

The Mathematical Messenger. Edited and Published by G. 
H. Harvill, A. B., Simsboro, La. Published bi-monthly. Price, 
$1.00 per year. 

This Journal is devoted entirely to Mathematics. The best mathemati- 
cians in the United States contribute to it. 

The Mathematical Magazine. Edited and Published by Arte- 
mas Martin, A. M., Ph. D., LL. D., Washington, D. C. Publised 
quarterly. Price, $1.00 per year. 

This Journal is devoted to the solution of problems in Arithmetic, Alge- 
bra, Geometry and Trigonometry. In it, are also published papers on 
various topics in Mathematics. 

The Mathematical Visitor. Edited and Published by Artemas 
Martin, A. M., Ph. D., LL. D., Washington, D. C. Published 
yearly. Price, $0.50. 

In this Journal are published solutions to the most difficult problems that 
have ever been solved The best Mathematicians from all over the world 
contribute to its pages. It is the most artistic Mathematical Journal pub- 
lished in America. 



